{"id":162088,"date":"2023-10-19T16:23:43","date_gmt":"2023-10-19T10:53:43","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=162088"},"modified":"2023-10-19T16:23:43","modified_gmt":"2023-10-19T10:53:43","slug":"ml-aggarwal-class-12-maths-solutions-section-c-chapter-3-ex-3-2","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-12-maths-solutions-section-c-chapter-3-ex-3-2\/","title":{"rendered":"ML Aggarwal Class 12 Maths Solutions Section C Chapter 3 Linear Programming Ex 3.2"},"content":{"rendered":"

Practicing ML Aggarwal Class 12 Solutions<\/a> Chapter 3 Linear Programming Ex 3.2 is the ultimate need for students who intend to score good marks in examinations.<\/p>\n

ML Aggarwal Class 12 Maths Solutions Section C Chapter 3 Linear Programming Ex 3.2<\/h2>\n

Solve the following (1 to 12) linear programming problems graphically :<\/p>\n

Question 1.
\nMinimize Z = – 3x + 4y subject to the constraints x + 2y \u2264 8, 3x + 2y \u2264 12, x \u2265 0, y \u2265 0.
\nAnswer:
\nDraw the straight lines x + 2y = 8
\nand 3x + 2y = 12
\nx, y \u2265 0
\neqn. (1) meet x-axis and y-axis at A (8, 0) and B (0, 4).
\neqn. (2) meet x-axis and y-axis at C (4, 0) and D (0, 6).
\neqn. (1) and eqn. (2) intersect at E (2, 3).
\n\"ML
\nThe shaded position OCEB is the feasible region and is bounded.
\nWe use corner point method to find the min. value of Z, where Z = – 3x + 4y<\/p>\n\n\n\n\n\n\n\n
Corner point<\/td>\nValue of Z<\/td>\n<\/tr>\n
O (0, 0)<\/td>\n0<\/td>\n<\/tr>\n
C (4, 0)<\/td>\n-3 \u00d7 4+ 4 \u00d7 0 = -12 (Min)<\/td>\n<\/tr>\n
E (2, 3)<\/td>\n-3 \u00d7 2 + 4 \u00d7 3 = 6<\/td>\n<\/tr>\n
B (0, 4)<\/td>\n-3 \u00d7 0 + 4 \u00d7 4 = 16<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u2234 Zmim<\/sub> = – 12 at corner point C (4, 0).<\/p>\n

Question 1(Old).
\nMaximize Z = 3x + 4y, subject to the constraints x + y \u2264 4, x \u2265 0, y \u2265 0.
\nAnswer:
\nGiven Max Z = 3x + 4y
\nSubject to constraints ; x + y \u2264 4
\nx, y \u2265 0
\nConverting the inequations into eqn\u2019s : x + y = 4
\n\"ML
\nFor region x + y < 4 ; The line x + y = 4
\nmeet coordinate axis at A (4, 0) and B (0, 4). Since (0, 0) lies on given inequation. Thus the region containing (0, 0) gives the solution set of given inequation.
\nThe shaded region OAB gives the feasible region and it is bounded. The corner points of feasible region are O (0, 0); A (4, 0) and B (0, 4).<\/p>\n

The value of objective function at these corner points is given below :<\/p>\n\n\n\n\n\n\n
Corner point<\/td>\nValue of objective function Z = 3x + 4y<\/p>\n

 <\/td>\n<\/tr>\n

O (0, 0)<\/td>\n0<\/td>\n<\/tr>\n
A (4,0)<\/td>\n12<\/td>\n<\/tr>\n
B (0, 4)<\/td>\n16 (Maximum)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u2234 Zmax<\/sub> = 16 at x = 0 and y = 4<\/p>\n

\"ML<\/p>\n

Question 2.
\nMinimize Z = 3x + 5y subject to x + 3y \u2265 3, x + 3y > 2, x \u2265 0, y \u2265 0.
\nAnswer:
\nDraw the straight lines x + y = 2 ………….(1)
\nand x + 3y = 3 …………..(2)
\nx, y \u2265 0
\neqn. (1) meet x-axis and .y-axis at A (2, 0) and B (0, 2).
\nand eqn. (2) meet x-axis and^-axis at C (3, 0) and D (0, 1).
\neqn. (1) and (2) intersects at E
\nThe shaded portion represents the feasible region determined by given constraints. Clearly the region CEB is feasible and unbounded.
\nWe calculate Z = 3x + 5y at corner points B (0, 2) ; C (3, 0) ; E (3\/2, 1\/2) :<\/p>\n\n\n\n\n\n\n
Corner point<\/td>\nValue of Z<\/td>\n<\/tr>\n
B(0, 2)<\/td>\n0 + 10 = 10<\/td>\n<\/tr>\n
C(3, 0)<\/td>\n9 + 0 = 9<\/td>\n<\/tr>\n
E(\\(\\frac{3}{2}\\), \\(\\frac{1}{2}\\))<\/td>\n\\( \\frac{9}{2}+\\frac{5}{2} \\) = 7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"ML
\nFrom the table, 7 be the smallest value of Z. But the feasible region is unbounded so 7 may or may not be minimum value of Z.
\nFor this, we draw 3x + 5y < 7. Since the open half plane has no common points with the feasible region.
\nSo the smallest value be the minimum value.
\n\u2234 Z = 3x + 5y has no max. value and Zmin<\/sub> = 7 at x = \\(\\frac{3}{2}\\) and y = \\(\\frac{1}{2}\\).<\/p>\n

Question 3.
\nMaximize Z = 4x + y, subject to the constraints x + y \u2264 50, 3x + y \u2264 90, x \u2265 0, y \u2265 0
\nAnswer:
\nDraw the straight lines x+y = 50 …………..(1)
\nand 3x + y = 90 …(2)
\nx, y \u2265 0
\neqn. (1) meets coordinates axis at A(50, 0) and B(0, 50)
\neqn. (2) meet x-axis and y-axis at C(30, 0) and D (0, 90).
\neqns. (1) and (2) intersects at E (20, 30).
\nNow x = 0 is y-axis and y = 0 is x-axis.
\nThe shaded portion represents the feasible region determined by given constraints.
\nClearly the region OCEB is feasible and bdd.
\nWe find max. value of Z using corner point method
\nwhere Z = 60x + 15
\n\"ML
\nThe coordinates of O, C, E and B are (0, 0), (30, 0), (20, 30) and (0, 50) respectively.<\/p>\n\n\n\n\n\n\n\n
Corner point<\/td>\nValue of Z<\/td>\n<\/tr>\n
O(0, 0)<\/td>\n60 \u00d7 0 + 15 \u00d7 0 = 0<\/td>\n<\/tr>\n
C(30, 0)<\/td>\n60 \u00d7 30 + 15 \u00d7 0 = 1800 (Max)<\/td>\n<\/tr>\n
E(20, 30)<\/td>\n60 \u00d7 20 + 15 \u00d7 30 = 1650<\/td>\n<\/tr>\n
B(0, 50)<\/td>\n60 \u00d7 0 + 15 \u00d7 50 = 750<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u2234 Zmax<\/sub> = 1800 at point C(30, 0)<\/p>\n

Question 3(Old).
\nMaximize Z = 3x + 2y subject to the constraints x + 2y \u2264 10, 3x + y \u2264 15, x \u2265 0, y \u2265 0
\nAnswer:
\nDraw the given st. lines
\nx + 2y = 10
\nand 3 x + y = 15
\nx, y \u2265 0
\neqn. (1) meet x axis and y-axis at A (10, 0) and B (0, 5)
\nand eqn. (2) meet x-axis andy-axis at C (5, 0) (ii) D (0, 15)
\nBoth eqn. (1) and (2) intersects at E (4, 3).
\nAlso x = 0 is y-axis and y = 0 is x-axis.
\nThus the shaded portion represents the feasible region determined by given constraints.
\nClearly OCEB is feasible and bounded.
\nso we calculate maf value of Z using corner point method, where Z = 3x + 2y
\nThe coordinates of O, C, E and B are (0, 0), (5, 0), (4, 3) and (0, 5) respectively.<\/p>\n\n\n\n\n\n\n\n
Corner point<\/td>\nValue of Z<\/td>\n<\/tr>\n
O (0, 0)<\/td>\n0<\/td>\n<\/tr>\n
C (5, 0)<\/td>\n3 \u00d7 5 + 1 \u00d7 0 = 15<\/td>\n<\/tr>\n
E (4, 3)<\/td>\n3 \u00d7 4 + 3 \u00d7 2 = 18<\/td>\n<\/tr>\n
B (0, 5)<\/td>\n3 \u00d7 0 + 2 \u00d7 5 = 10<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Max. Z = 18 at point E (4, 3).
\n\"ML<\/p>\n

\"ML<\/p>\n

Question 4.
\nMinimize Z = 200x + 500p, subject to the constraints x + 2y \u2265 10, 3x + 4y \u2264 24, x \u2265 0, y \u2265 0.
\nAnswer:
\nDraw the straight lines x + 2y = 10 …(1) ‘
\nand 3x + 4y = 24 …(2) and x, y \u2265 0
\neqn. (1) meets coordinate axes at A(10,0) and B(0, 5).
\neqn. (2) meets coordinate axes at C(8, 0) and D (0, 6).
\neqn. (1) and (2) intersects at E (4, 3).
\n\"ML
\nThe shaded position BED is the feasible region and is bounded.
\nTo determine min. value of Z, we use corner point method.
\nwhere Z = 200x + 500y<\/p>\n\n\n\n\n\n\n
Corner point<\/td>\nZ<\/td>\n<\/tr>\n
B (0, 5)<\/td>\n200 x 0 + 500 x 5 = 2500<\/td>\n<\/tr>\n
E (4, 3)<\/td>\n200 x 4 + 500 x 3 = 2300 (Min)<\/td>\n<\/tr>\n
D (0, 6)<\/td>\n200 x 0 + 500 x 6 = 3000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u2234 Min. Z = 2300 at E (4, 3).<\/p>\n

Question 5.
\nMinimize Z = x + 2y subject to 2x + y \u2265 3, x + 2y \u2265 6, x \u2265 0, y \u2265 0. Show that the minimum value of Z occurs at more than two points.
\nAnswer:
\nConverting the given inequations into equations :
\n2x + y = 3
\nx + 2y = 6
\nx = y = 0
\nThe region 2x + y > 3 : The line 2x + y = 3 meets coordinate axis at A (0, 0) does not lies on 2x + y > 3. Thus, the region not containing (0, 0) gives the solution set of given inequation.<\/p>\n

Region x + 2y > 6 ; The line x + 2y = 6 meets coordinate axis at C (6, 0) and D (0, 3). Since (0, 0) does not lies on given region and hence region not containing (0, 0) gives the solution set of given inequation.
\nRegion x > 0, y > 0 ; It represents the first quadrant of XOY plane.
\nAlso both lines 2x + y – 3 and x + 2y = 6 intersects at P (0, 3)
\nSo the shaded area CP gives the feasible region and it is unbounded with corner points C (6,0) and P (0, 3).<\/p>\n\n\n\n\n\n
Corner point<\/td>\nZ = x + 2y<\/td>\n<\/tr>\n
C (6, 0)<\/td>\n6<\/td>\n<\/tr>\n
P (0, 3)<\/td>\n6<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"ML
\nZ has minimum value at C (6, 0) on the line and P (0, 3) i.e. at all points on the line joining C and P. Thus Z has minimum value occurs at more than two points.<\/p>\n

Question 6.
\nMinimize and maximize Z = 3x + 9y subject to the constraints x + y \u2265 10, x + 3y \u2264 60, x \u2264 y, x \u2265 0, y \u2265 0. (NCERT)
\nAnswer:
\nDraw the straight lines x + 3y = 60 …(1)
\nx + y = 10 …(2)
\nand x = y ..(3)
\neqn. (1) meet x-axis and y-axis at A (60, 0) and B (0, 20).
\neqn. (2) meet x-axis and y-axis at C(10, 0) and D (0, 10)
\neqn. (3) passes through (0, 0).
\neqn. (3) meets eqn. (1) at M (15, 15) and meets eqn. (2) at N(5, 5).
\nThe shaded portion represents the feasible region determined by given constraints.
\nAlso eqn. (1) and (2) intersects at P (- 15, 25).
\nClearly the feasible region DNMB is bounded.
\nWe use corner point method to find the max. and min value of Z where Z = 3x + 9y<\/p>\n\n\n\n\n\n\n\n
Corner point<\/td>\nValue of Z<\/td>\n<\/tr>\n
D(0, 10)<\/td>\n3 \u00d7 0 + 9 \u00d7 10 = 90<\/td>\n<\/tr>\n
N (5, 5)<\/td>\n3 \u00d7 5 + 9 \u00d7 5 = 60 (Min)<\/td>\n<\/tr>\n
M (15, 15)<\/td>\n3 \u00d7 15 + 9 \u00d7 15 = 180<\/td>\n<\/tr>\n
B (0, 20)<\/td>\n3 \u00d7 0 + 9 \u00d7 20 = 180 (Max)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"ML
\n\u2234 Min Z = 60 at point (5, 5) and Max. Z = 180 at the points (15, 15) and (0, 20).
\ni.e. Maximum value of Z occurs at all points on the line joining M and B.<\/p>\n

Question 7.
\nMinimize Z = 3x + 2y subject to the constraints x + y \u2265 8, 3x + 5y \u2264 15, x \u2265 0, y \u2265 0. (NCERT)
\nAnswer:
\nConverting the given inequations into equations :
\n-x + y = 8 ;
\n3x + 5y= 15 ;
\nx = 0 = y
\nFor region x + y \u2265 8 ; The line x + y = 8 meets the coordinate axis at A (8, 0) and B (0, 8). Since (0, 0) does not lies on x + y \u2265 8.
\nThus, region not containing (0, 0) gives the soln. set of given inequation.
\nFor Region 3x + 5y \u2264 15 ;
\nThe line 3x + 5y = 15 meets coordinate axis at C (5, 0) and D (0, 3).
\nSince (0, 0) lies on 3x + 5y \u2264 15.
\nThus the region containing (0, 0) gives the solution set of given inequation.
\nFor region x, y \u2265 0 ; It represents the first quadrant of xoy plane.
\nAlso both lines intersects at P\\(\\left(\\frac{25}{2},-\\frac{9}{2}\\right)\\)
\nClearly their is no common feasible region.
\n\"ML<\/p>\n

Question 8.
\nMinimize and mximize Z = x + 2y subject to the constraints
\nx + 2y \u2265 100, 2x – y \u2264 0, 2x + y \u2264 200, x \u2265 0, y \u2265 0. (NCERT)
\nAnswer:
\nDraw the st. lines x + 2y= 100 …(1)
\n2x – y = 0 …(2)
\nand 2x + y = 200 ……….(3)
\nx, y \u2265 0
\neqn. (1) intersects x-axis and 7-axis at A (100, 0) and B (0, 50)
\n(2) line meets x-axis and 7-axis at C (100, 0) and D (0, 200)
\neqn. (2) meet eqn. (1) at M (20, 40) and meet eqn. (3) at N (50, 100). eqn. (1) and (3) intersects at point P (100, 0).
\nThe shaded region BMND is feasible and bounded we determine the max or min. value of Z = x + 2y using corner point method.
\nZ = x + 2y<\/p>\n\n\n\n\n\n\n\n
Corner point<\/td>\nZ = x + 2y<\/td>\n<\/tr>\n
B(0, 50)<\/td>\n0 + 2 \u00d7 50 = 100<\/td>\n<\/tr>\n
M(20, 40)<\/td>\n20 + 2 \u00d7 40 = 100<\/td>\n<\/tr>\n
N(50, 100)<\/td>\n50 + 2 \u00d7 100 = 250<\/td>\n<\/tr>\n
D(0, 200)<\/td>\n0 + 2 \u00d7 200 = 400 (Max)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"ML
\n\u2234 Max. Z = 400 at the point (0, 200).
\nand Min. Z = 100 at the points (0, 50) and (20, 40) i.e. line segment joining B and M.<\/p>\n

Question 9.
\nMaximize Z = x + y subject to the constraints x – y \u2264 -1, -x + y \u2264 0, x \u2265 0, y \u2265 0. (NCERT)
\nAnswer:
\nWe draw the straight lines x – y = – 1 …(1)
\nand -x + y = 0 ……….(2)
\nx, y \u2265 0
\nline (1) meet x-axis andy-axis at A (- 1, 0) and B (0, 1)
\nline (2) meet x-axis and y-axis at C (0, 0)
\n(1) and (2) intersects at no point.
\nClearly there is no point which satisfies all constraints simultaneously.
\nThus there is no feasible region and hence there is no feasible solution.
\n\"ML
\nHence maximum value of Z does not exists.<\/p>\n

\"ML<\/p>\n

Question 10.
\nA manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine a and 1 hour on machine B to produce a package of bolts. He earns a profit of \u20b9 17.50 per package on nuts and \u20b9 7 per package on bolts. How many packages of each should be produced each day, so as to maximize his profit, if he operates his machines for atmost 12 hours a day.
\nAnswer:
\nThe given information can be summarised in tabular form is given as under :<\/p>\n\n\n\n\n\n\n\n
Time required to produce products<\/td>\n<\/tr>\n
Machines<\/td>\nNuts<\/td>\nBolts<\/td>\nMax. machine hours available<\/td>\n<\/tr>\n
A<\/td>\n1<\/td>\n3<\/td>\n12<\/td>\n<\/tr>\n
B<\/td>\n3<\/td>\n1<\/td>\n12<\/td>\n<\/tr>\n
Profit (in \u20b9)<\/td>\n\u20b9 17.50<\/td>\n\u20b9 7<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Let the manufacturer produces x-package of nuts and y package of bolts each day. Since machine A takes 1 hour to produce one package of nuts and 3 hours to produce one package of bolts and max. time available for machine A is 12 hours.
\nThus, x + 3y \u2264 12
\nSimilarly for machine, the constraint be given as; 3x +y \u2264 12
\nNow it is given that, profit of \u20b9 17.50 per package on nuts and \u20b9 7 per package of bolts.
\nThus total profit on x-package of nuts and y-package of bolt be 17.50x + 7y.
\nLet Z be the total profit. Then Z = 17.50x + 7y.
\nClearly x \u2265 0 and y \u2265 0
\nThus the given LPP can be stated mathematically as;
\nx + 3y \u2264 12
\n3 x + y \u2264 12
\nx \u2265 0, y \u2265 0
\nand Max Z = 17.50x + 7y
\nFor region x + 3y< 12; The line x + 3y = 12 meets coordinate axes at A (12, 0) and B (0, 4). Clearly (0,0) lies on x + 3y \u2264 12 [\u2235 0 \u2264 12] Thus, region containing (0, 0) gives the soln. set of x + 3y \u2264 12 For region 3x + y \u2264 12; The line 3x + y = 12 meets coordinate axes at C (4, 0) and D (0, 12). Also (0, 0) lies on inequality. Thus region containing (0, 0) gives the solution set of given inequality. Clearly x > 0, y > 0 represents the first quadrant of XOY plane.
\nThe lines x + 3y = 12 and 3x + y = 12 intersect at P (3, 3).
\nNow the bounded shaded region OCPBO represents the feasible region with comer points 0(0,0) ; C(4, 0) ; P(3, 3) and B(0, 4).
\n\"ML
\nThe values of objective function at these comer points is given as under :<\/p>\n\n\n\n\n\n\n\n
Corner points<\/td>\nZ = 17.50x + 7y<\/td>\n<\/tr>\n
O(0, 0)<\/td>\nZ = 0<\/td>\n<\/tr>\n
C (4, 0)<\/td>\nZ = 17.50 \u00d7 4 + 7 \u00d7 0 = 70<\/td>\n<\/tr>\n
P (3, 3)<\/td>\nZ = 17.50 \u00d7 3 + 7 \u00d7 3 = 73.50<\/td>\n<\/tr>\n
B (0, 4)<\/td>\nZ = 0 + 7 \u00d7 4 = 28<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Clearly Z is maximum at x = y = 3 and Zmax<\/sub> = \u20b9 73.50
\nHere the manufacturer produces 3 packages of each nuts and bolts per day to realise maximum profit of \u20b9 73.50.<\/p>\n

Question 10.
\nMaximize Z = -x + 2y subject to the constraints x \u2265 3, x + y \u2265 5, x + 2y \u2265 6, y \u2265 0. (NCERT)
\nAnswer:
\nConverting the given inequations into equations. x = 3 ;x + y = 5 ; x + 2y = 6 ; y = 0
\nFor region x \u2265 3 : The line x = 3 is a line passing through A (3, 0) parallel to y-axis. Since (0, 0) does not lies on x \u2265 3. So region not containing (0, 0) gives the solution set of x > 3.
\nFor region x + y = 5 ; The line x + y = 5 meets coordinate axis at B (5, 0) and C (0, 5). Since (0, 0) does not lies on x + y \u2265 5. Thus region not containing (0, 0) gives the solution set of given inequation.
\nFor region x + 2y \u2265 6 : The line x + 2y = 6 meets coordinate axis, at D (6, 0) and E (0, 3). Since (0, 0) does not lies on x +2y \u2265 6. Thus, the region not containing origin gives the soln. set of given inequation.
\nRegion x, y \u2265 0 represents the first quadrant of XOY plane.
\nThe lines x = 3 and x + y = 5 intersects at P (3, 2).
\nThe lines x = 3 and x + 2y = 6 intersects at Q(3, \\(\\frac{3}{2}\\)).
\nThe lines x + y = 5 and x + 2y = 6 intersects at R (4, 1).
\nThe shaded position PRD gives the feasible region.
\nSo min value of Z exists but maximum value of Z does not exists since feasible region is unbounded.
\n\"ML<\/p>\n

\"ML<\/p>\n

Question 11.
\nA small firm manufactures gold rings and chains. The combined number of rings and chains manufactured per day is atmost 24. It takes one hour to make a ring and half an hour for a chain. The maximum number of hours available per day is 16. If the profit on a ring is ? 300 and on a chain is ? 190, how many of each should be manufactured daily so as to maximize the profit ?
\nAnswer:
\nLet the required number of gold rings and chains manufactured by a firm are x and y respectively to get the maximum profit.
\nSince, profit on each ring and chains are \u20b9 300 and \u20b9 190 respectively, so, profit onx units of rings and y units of chains are \u20b9 300x and \u20b9 190y respectively
\nLet Z be total profit. Then Z = 300x + 190y and we have to maximize.
\nSince each unit of ring and chain require 1 hr and 30 min to make respectively, so, x units of rings and y units of rings require 60x and 30y min. to make respectively, but given total time available to make it, is 16 \u00d7 60 = 960, so (first constraint) is given by .
\n60x + 30y \u2264 960 Thus, 2x + y \u2264 32
\nGiven, total number or rings and chains manufactured is at most 24, so, second constraint is given by x + y \u2264 24
\nHence, the mathematical formulation of given LPP is given below:
\nMaximum Z = 300x + 160y
\nSubject to constraints,
\n2x + y \u2264 32 x + y \u2264 24
\nx, y \u2265 0 [Since production can not be less than zero]
\nRegion 2x +y \u2264 32 : The line 2x +y = 32 meets coordinate axes at C( 16,0) & D(0, 32) respectively. Region containing origin represents the solution set of 2x + y \u2264 32 as (0,0) satisfies 2 x + y \u2264 32. Region x + y \u2264 24 : line x + y = 24 meets coordinate axes at A(24, 0) & B(0, 24) respectively. Region containing origin represents the solution set of x + y \u2264 24 as (0,0) satisfies x + y \u2264 24. Region x, y \u2265 0 : it represents the first quadrant of xy-plane.
\nThus Shaded region OCEB represents feasible region.
\n\"ML
\nThe point E(8, 16) is obtained by solving 2 x+y = 32 and x +y = 24 simultaneously.<\/p>\n\n\n\n\n\n\n\n
Corner Point<\/td>\nZ = 300x + 160y<\/td>\n<\/tr>\n
0(0,0)<\/td>\n300(0) + 160(0) = 0<\/td>\n<\/tr>\n
C(16,0)<\/td>\n300(16) + 160(0) = 4800<\/td>\n<\/tr>\n
E(8, 16)<\/td>\n300(8) + 160(16) = 4960<\/td>\n<\/tr>\n
B(0,24)<\/td>\n300(0) + 160(24) = 3840<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Zmax<\/sub> or maximum Z = 4960 at x = 8, y = 16
\nreqd. number of rings = 8 & reqd. no. of chains = 16 & maximum profit = \u20b9 4960<\/p>\n

Question 12.
\nA company produces two types of goods, A and B, that require gold and silver. Each unit of type A requires 3 g of silver 1 g of gold wyhile that of B requires 1 g of silver and 2 g of gold. The company can use at most 9 g of silver and 8 g of gold. If each unit of type A brings a profit of Rs. 40 and that of types B Rs. 50, find the number of units of each type that the company should produce to maximize the profit. Formulate and solve graphically the LPP and find the maximum profit.
\nAnswer:
\nLet x be the number of units of good A and y be the no. of units of good B produce d by the company. So total profit of company on x units of good A and y units of good B be 40x + 50y.
\nLet Z be the total profit of company.
\nThen Z = 40x + 50y
\nWe make the given data in tabulated form as under :<\/p>\n\n\n\n\n\n\n
goods<\/td>\nsilver (in gm)<\/td>\ngold (in gm)<\/td>\nRequirement<\/td>\n<\/tr>\n
A<\/td>\n3<\/td>\n1<\/td>\n<\/td>\n<\/tr>\n
B<\/td>\n1<\/td>\n2<\/td>\n<\/td>\n<\/tr>\n
Requirement<\/td>\n9<\/td>\n8<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

since it is given that, the company can use atmost 9 gm of silver \u2234 3x + y < 9
\nAlso, it is given that, the company can use atmost 8 gm of gold \u2234 x + 2y < 8
\nThus the mathematical modeling of given LPP is as under :
\nMax Z = 40x + 50y
\nSubject to constraints : 3x + y \u2264 9; x + 2y \u2264 8 ; x \u2265 0, y \u2265 0 [since no. of units of good A and B can\u2019t be negative]
\nFor region 3x + y \u2264 9; The line 3x + y = 9 meets coordinate axes at A(3, 0) and B(0, 9). Clearly (0, 0) satisfies the given inequality. Thus region containing (0, 0) gives the solution set of given inequality.
\nFor region x + 2y \u2264 8; The line x + 2y = 8 meets coordinate axes at C(8, 0) and D(0, 4). Clearly (0, 0) satisfies the given inequality. .
\nThus, region containing (0, 0) gives the solution set of given inequality.
\nRegion x \u2265 0, y \u2265 0 represents the first quadrant of XOY plane. Both lines 3x + y = 9 and x + 2y = 8 intersects at P(2, 3).
\nThus the shaded region OAPD represents the bounded feasible region and its comer points are 0(0, 0); A(3, 0); P(2, 3) and D(0, 4).
\nWe evaluate the objective function Z at these comer points.<\/p>\n\n\n\n\n\n\n\n
Corner points<\/td>\nZ = 40x + 50y<\/td>\n<\/tr>\n
O(0, 0)<\/td>\n0<\/td>\n<\/tr>\n
A(3, 0)<\/td>\n40 \u00d7 3 + 50 \u00d7 0 = 120<\/td>\n<\/tr>\n
P(2, 3)<\/td>\n40 \u00d7 2 + 50 \u00d7 3 = 230<\/td>\n<\/tr>\n
D(0, 4)<\/td>\n40 \u00d7 0 + 50 \u00d7 4 = 200<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Clearly Zmax<\/sub> = 230 at P(2, 3) i.e. x = 2 and y = 3
\nThus, the company should produce 2 units of good A and 3 units of good B to get maximum profit Rs. 230.<\/p>\n

\"ML<\/p>\n

Question 13.
\nA factory manufactures two types of screws, A and B, each type requiring the use of two machines – an automatic and a hand-operated. It takes 4 minutes on the automatic and 6 , minutes on the hand-operated machine to manufacture a package of screws \u2018A\u2019, while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a package of screws \u2018B’ Each machine is available for atmost 4 hours on any day. The manufacturer can sell a package of screws \u2018A\u2019 at a profit of \u20b9 7 and of screws \u2018B\u2019 at a profit of \u20b9 10. Assuming that he can sell all the screws he can manufacture, how many\u2003packages of each type should the factory owner produce in a day in order to maximize his profit ? Determine the maximum profit. (NCERT)
\nAnswer:
\nLet the factory manufactures x screws of type A and y screws of type B on each day to maxmise the profit.
\nTherefore, x \u2265 0 and y \u2265 0
\n\"ML
\nThe given information can be tabulated as given below<\/p>\n\n\n\n\n\n
<\/td>\nScrew A<\/td>\nScrew B<\/td>\nAvailability<\/td>\n<\/tr>\n
Automatic Machine (min)<\/td>\n4<\/td>\n5<\/td>\n4 \u00d7 60 = 240<\/td>\n<\/tr>\n
Hand Operated Machine (min)<\/td>\n6<\/td>\n3<\/td>\n4 \u00d7 60 = 240<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The profit on a package of screws A is \u20b9 7 and on the pakage of screws B is \u20b9 10. Therefore, the constraints are;
\n4x + 6y \u2264 240
\n6x + 3y \u2264 240
\nTotal profit, Z = 7x + 10y
\nThe required mathematical formulation of the given problem is as follows :
\nMaximum Z = 7x + 10y Subject to the constraints,
\n4x + 6y \u2264 240
\n6x + 3y \u2264 240
\nx, y \u2265 0
\nConverting given inequations into equations;
\n4x + 6y = 240 \u21d2 2x + 3y = 120 …(1)
\n& 6x + 3y = 240 \u21d2 2x +y = 80 ……..(2)
\nand x = y = 0
\nFor Region 2x + 3y \u2264 120 : The line 2x + 3y = 120 meets coordinate axes at A1<\/sub> (60, 0) and B1<\/sub> (0, 40). Also (0,0) satisfies 2x + 3y \u2264 120. Thus the region containing (0, 0) gives the solution set of 2x + 3y \u2264 120.
\nFor Region 2x + y \u2264 80 : The line 2x + y = 80 meets coordinate axes at A2<\/sub> (40, 0) and B2<\/sub> (0, 80) Clearly (0,0) satisfies it. These the region containing (0, 0) gives the solution set of inequation 2x + y \u2264 80.
\nRegion x \u2265 0; y \u2265 0 represents the first quadrant, both lines (1) & (2) intersects at P (30, 20).
\nThe feasible region determined by the system of constraints is given by shaded area given below : The comer points are 0(0,0), A2(40, 0), P (30, 20) and Bj (0, 40).
\nThe values of Z at these corner points are as follows :<\/p>\n\n\n\n\n\n\n\n
Corner point<\/td>\nZ = 7x + 10y<\/td>\n<\/tr>\n
O (0,0)<\/td>\n0<\/td>\n<\/tr>\n
A2(40,0)<\/td>\n280<\/td>\n<\/tr>\n
P (30,20)<\/td>\n410 (Maximum)<\/td>\n<\/tr>\n
B1 (0, 40)<\/td>\n400<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The maximum value of Z is 410 i.e. Zmax<\/sub> = 410 at (30, 20).
\nHence, the factory should produce 30 package of screws A and 20 packages of screws B to get the maximum profit of \u20b9 410.<\/p>\n

Question 13(Old).
\nA manufacturer considers that men and women workers are equally efficient and so he ^ pays them at the same rate. He has 30 and 17 units of workers (male and female) and 0^ capital respectively, which he uses to produce two types of goods A and B. To produce one unit of A, 2 workers and 3 units of capital are required while 3 workers and 1 unit of capital is required to produce one unit of B. If A and B are priced at ? 100 and ? 120 per unit respectively, how should he use his resources to maximise the total revenue ? Form the above as an L.P.P. and solve graphically.
\nDo you agree with this view of the manufacturer that men and women workers are equally efficient and so should be paid at the same rate ?
\nAnswer:
\nLet x units of workers and y units of capital are required to maximise the total revenue. Then the mathematical model of the LPP is given as follows :
\nMaximize Z = 100x + 120y Subject to constriants ;
\n2x + 3y \u2264 30
\n3x + y \u2264 17
\nand x \u2265 0, y \u2265 0
\nTo solve the LPP we convert the inequations into equations :
\n2x + 3 y = 30
\nx + y = 17
\nx = 0 = y
\nFor region 2x + 3y \u2264 30 ; The line 2x + 3y = 30 meets the coordinate axes at A1<\/sub> (15, 0) and A (0, 10). Clearly (0, 0) satisfies the inequation 2x + 3y \u2264 30. Hence the region containing (0, 0) gives the solution set of given inequation.
\nFor region 3x + y \u2264 17 ; The line 3x + y = 17 meets coordinate axes at A2<\/sub> (17\/3, 0) and B2<\/sub> (0, 17). Clearly (0, 0) satisfies 3x + y \u2264 17. Hence the region containing (0, 0) gives the solution set of given inequation.
\nFurther, x \u2265 0, y \u2265 0 represents the first quadrant of xoy plane.
\nBoth lines 2x + 3y = 30 and 3x + y = 17 intersects at C(3, 8).
\nThe shaded area gives the feasible region OA2<\/sub>PB1<\/sub>
\n\"ML
\nThe coordinates of the vertices (Corner – points) of shaded feasible region OA2<\/sub>CA are O (0, 0), A2<\/sub>(\\(\\frac{17}{3}\\), 0), C(3, 8) and A(0, 10)
\nThe values of the objective of function at these comer points are given in the following table :<\/p>\n\n\n\n\n\n\n\n
Corner Point<\/td>\nValue of objective function
\nZ = 100.x + 120y<\/td>\n<\/tr>\n
A2<\/sub>(\\(\\frac{17}{3}\\), 0)<\/td>\nZ = 100 \u00d7 y+ 120 \u00d7 0 = 566. 67<\/td>\n<\/tr>\n
C(3, 8)<\/td>\nZ = 100 \u00d7 3 + 120 \u00d7 10 = 1260<\/td>\n<\/tr>\n
A(0, 10)<\/td>\nZ = 100 \u00d7 0 + 120 \u00d7 10 = 1200<\/td>\n<\/tr>\n
O(0, 0)<\/td>\nZ = 0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Clearly Zmax<\/sub> = 1260 at x = 3 and y = 8
\nHence, 3 units of workers and 8 units of capital must be used to maximize the profit. Thus, the required maximum profit that can be earned is \u20b9 1260.
\nYes, because efficiency of a person does not depend on sex (male or female).<\/p>\n

\"ML<\/p>\n

Question 14.
\nA dealer in rural area wishes to purchase a number of sewing machines. He has only \u20b9 57600 to invest and has space for atmost 20 items. An electronic sewing machine costs him \u20b9 3600 and a manually operated sewing machine \u20b9 2400. He can sell an electronic sewing machine at a profit of \u20b9 220 and a manually operated sewing machine at a profit of \u20b9 180. Assuming that he can sell all the items that he can buy, how should he invest his money in order to maximize his profit. Make it as an L.P.P. and solve it graphically.
\nAnswer:
\nLet x and y are the number of electronic and manually operated sewing machine that a dealer buys and sells.
\nThe dealer gets a profit of \u20b9 220 and \u20b9 180 on selling on electronic and manually operated sewing machine. .
\nSo Total profit of dealer = 220x + 180y
\nLet Z be the total profit of dealer.
\nSince cost of 1 electronic and 1 manually operate sewing machine are \u20b9 3600 and \u20b9 2400 respectively.
\nThus cost of x electronic and y manually operated sewing machines are 3600x and 2400y. also the dealer can investment atmost \u20b9 57600.
\nThus purchased constraint is given by 3600x + 2400y \u2264 57600 \u21d2 3x + 2y \u2264 48
\nAlso the dealer has a space to store atmost 20items.
\nThus storage constraint is given by x + y \u2264 20
\nHence, the mathematical formulation of given LPP is given as under:
\nMax Z = 220x + 180
\nSubject to Constraints ; 3x + 2y \u2264 48
\nx + y \u2264 20
\nx \u2265 0 ; y \u2265 0
\n[Since electronic and manually operated sewing machines can\u2019t be negative]
\nTo solve LPP, we convert inequations into eqn\u2019s 3x + 2y = 48; x + y = 20; x = 0 = y
\nFor region 3x + 2y \u2264 48 ; The line 3x + 2y = 48 meets coordinate axis at A (16, 0) and B (0, 24).
\nSince (0, 0) lies on given inequation. Thus, region containing (0, 0) gives soln. set of 3x + 3y \u2264 48. For region x + y \u2264 20 ;
\nThe line x + y = 20 meets coordinate axis at C (20, 0) and D (0, 20).
\nSo region containing (0, 0) gives the soln. set of given inequation since (0, 0) lies on x + y \u2264 20.
\nThe region x \u2265 0, y \u2265 0 represents the first quadrant of XOY plane.
\nBoth lines intersects at P (8, 12).
\nThus the shaded region OAPD gives the bounded feasible region with corner points O (0, 0); A (16, 0) ; P (8, 12) and D (0, 20).
\nWe evaluate Z at these corner points<\/p>\n\n\n\n\n\n\n\n
Corner point<\/td>\nZ = 220x + 180y<\/td>\n<\/tr>\n
O (0, 0)<\/td>\n0<\/td>\n<\/tr>\n
A (16, 0)<\/td>\n220 \u00d7 16 = 3520<\/td>\n<\/tr>\n
P(8, 12)<\/td>\n220 \u00d7 8 + 180 \u00d7 12 = 3920 (Max)<\/td>\n<\/tr>\n
D(0, 20)<\/td>\n220 \u00d7 0 + 180 \u00d7 20 = 3600<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"ML
\n\u2234 Zmax<\/sub> = 3920 at P (8, 12) i.e. at x = 8 ; y = 20
\nrequired no. of electronic sewing machines = 8
\nrequired no. of normally operated machines = 12
\nand Max. profit = \u20b9 3920
\nKeeping in view of saving environment and conservation of exhausitible resources, the manually operated machines should be promoted.<\/p>\n

Question 14 (Old).
\nA company produces two types of items P and Q. Manufacturing of both items requires the metals gold and copper. Each unit of item P requires 3 gms of gold and 1 gm of copper while N that of item Q requires 1 gm of gold and 2 gms of copper. The company has 9 gms of gold and 8 gms of copper in store. If each unit of item P makes a profit of \u20b9 50 and each unit of item Q makes a profit of \u20b9 60, determine the number of units of each item that the company should produce to maximize profit. What is the maximum profit? (ISC 2010)
\nAnswer:
\nLet number of item P and Q be x and y respectively are required to maximise the profit. Since, profits on each P and Q are \u20b9 50 and \u20b9 60 respectively. So, profits on x goods of type P and y goods of type Q are 50x and 60y respectively,
\nLet Z be total profit on A and B.
\nThen Z = 50x + 60y
\nSince, each P and Q require 3 gm and 1 gm of gold respectively. So, x items of type P and y items of type Q require 3x and y gm gold respectively but, given total gold available is 9gm. So, (first constraint) is given by
\n3x + y \u2264 9
\nSince, each item of type P and Q require 1 gm and 2gm of copper respectively, so, x item of type P and y item of type Q require x gm and 2y gm of copper respectively but, given total copper available is 8gm. Then (second constraint) is given by
\nx + 2y \u2264 8
\nHence mathematical formulation of LPP is, given below:
\nMax Z = 50x + 60y
\nSubject to constraints,
\n3 x + y \u2264 9
\nx + 2y \u2264 8
\nx, y \u2265 0 [Since products of A and B can not be less than zero]
\nRegion 3x + y \u2264 9 : Then line 3x + y = 9 meets coordinate axes at A1<\/sub>(3, 0) & B1<\/sub>(0,9) respectively.
\nRegion containing origin respresents the solution set of 3x + y \u2264 9 as (0,0) satisfies 3x + y \u2265 9.
\nRegion x + 2y \u2264 8 : line x + 2y = 8 meets coordinate axes at C1<\/sub>(8,0) & D,(0,4) respectively.
\nRegion containing origin represents the solution set of x + 2y \u2264 8 as (0,0) satisfies x + 2y \u2264 8.
\nRegion x, y \u2265 0 : it represents first quardant of xy region.
\nShaded region OA1<\/sub>E1<\/sub> is the feasible region.
\nThe points E1<\/sub>(2,3) is obtained by solving 3x + y = 9 and x + 2y = 8 simultaneously.
\nThe corner points of feasible region are O(0, 0); A1<\/sub>(3, 0); E1<\/sub>(2, 3) and D1<\/sub>(0, 40).<\/p>\n\n\n\n\n\n\n\n
Corner point<\/td>\nZ = 50x + 60y<\/td>\n<\/tr>\n
O(0,0)<\/td>\n50(0) + 60(0) = 0<\/td>\n<\/tr>\n
A(3,0)<\/td>\n50(3) + 60(0)= 150<\/td>\n<\/tr>\n
E1<\/sub>(2,3)<\/td>\n50(2) + 60(3) = 280<\/td>\n<\/tr>\n
D1<\/sub>(0, 40)<\/td>\n50(0) + 60(4) = 240<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Therefore maximum Z = 280 at x = 2, y = 3
\n\"ML
\nHence, reqd. Maximum profit = \u20b9 280 and required number of goods of type A = 2, and reqd no. of goods of type B = 3.<\/p>\n

\"ML<\/p>\n

 <\/p>\n

Question 15.
\nA company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours available for assembling. The profit is \u20b9 5 each for type A and \u20b9 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize the profit ? (NCERT)
\nAnswer:
\nLet the company manufacture x souvenirs of type A and y souvenirs of type B.
\nTherefore, x \u2265 0 and y \u2265 0
\nThe given information can be put in tabular form is follows as :<\/p>\n\n\n\n\n\n
<\/td>\nType A<\/td>\nType B<\/td>\nAvailability<\/td>\n<\/tr>\n
Cutting (min)<\/td>\n5<\/td>\n8<\/td>\n3 x 60 + 20 = 200<\/td>\n<\/tr>\n
Assembing (min)<\/td>\n10<\/td>\n8<\/td>\n4 x 60 = 240<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The profit on type A souvenirs is \u20b9 5 and on type B souvanirs is \u20b9 6.
\nTotal profit, Z = 5x + 6y
\nThe required mathematical formulation of the given problem is Maximise Z = 5x + 6y
\nsubject to the constraints,
\n5x + 8y \u2264 200
\n5x + 4y \u2264 120
\nx, y \u2265 0
\nFor solution of LPP, We convert inequations into equations
\n5x + 8y = 200
\n& 5x + 4y = 120
\n& x = 0 = y
\n\"ML
\nFor region 5x + 8y \u2264 200 : The line 5x + 8y = 200 meet coordinate axes at A1<\/sub>(40,0) and B1<\/sub>(0, 25). so region containing (0,0) be the solution set of inequation 5x + 8y < 200 as (0,0) lies on 5x + 8y \u2264 200. For region 5x + 4y \u2264 120 : The line 5x + 4y = 120 meet coordinate axes at A2<\/sub>(24, 0) and B2<\/sub>(0, 30) and region contaning (0,0) represents the solution set of inequation 5x + 4y \u2264 120 as (0, 0) satisfies the inequation.Now region x \u2265 0; y \u2265 0 gives the first quadrant of xy plane.
\nBoth lines intersects at P (8, 20).
\nThe feasible region OA2<\/sub>PB1<\/sub> determined by the system of constraints is as follows.
\nThe corner points are O(0,0), A2<\/sub> (24,0) P (8, 20) and B1<\/sub> (0, 25).
\nThe values of Z at these corner points are as follows.<\/p>\n\n\n\n\n\n\n\n
Corner point<\/td>\nZ = 5x + 6y<\/td>\n<\/tr>\n
O(0,0)<\/td>\n0<\/td>\n<\/tr>\n
A2(24, 0)<\/td>\n= 5 \u00d7 24 + 6 \u00d7 0=120<\/td>\n<\/tr>\n
P(8, 20)<\/td>\n= 5 \u00d7 8 + 6 \u00d7 20 =160<\/td>\n<\/tr>\n
B1(0, 25)<\/td>\n150 – Maximum<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The maximum value of Z is 160 at (8, 20) i.e. x = 8 and y = 20
\nThus, 8 souvenirs of type A and 20souvenirs of type B should be produced each day to get the maximum profit of \u20b9 160.<\/p>\n

Question 15 (Old).
\nA cooperative society of farmers has 50 hactare of land to grow two crops X and Y. The Pr\u00b0fit from crop X and Y per hectare are estimated as \u20b9 10500 and \u20b9 9000 respectively. To control weeds, a liquid herbicide has to be used for crops X and Y at the rate of 20 litres and 10 litres per hectare. Further, no more than 800 litres of herbicide should be used in order to protect fish and wild life using a pond which collects drainage from this land.
\nKeeping in mind that the protection of fish and other wildlife is more important than earning profit, how much land be allocated to each crop so as to maximize the total profit ? Form an L.P.P. from the above and solve it graphically. Do you agree with the message that the protection of wildlife is utmost necessary to preserve the balance in environment ?
\nAnswer:
\nLet x hectares of land grows crop X. and y hectares of land grows crop Y.
\nThen the mathematical modelling of the LPP is given as follows :
\nMaximize Z = 10,500x + 9,000y
\nSubject to constraints ; x + y \u2264 50,
\n20x + 10y \u2264 800
\nand x \u2265 0, y \u2265 0
\nTo solve the LPP we convert inequation into equations x + y = 50
\n20x + 10y = 800 ; x = y = 0
\nFor region x + y \u2264 50 ; The line x + y = 50 meets coordinate axes at A1<\/sub>(50, 0) and B1<\/sub> (0, 50).
\nSince (0, 0) satisfies given in equation so the region containing (0, 0) gives the solution set of x + y < 50.
\nFor region 20x + 10y < 800 ;
\nThe line 20x + 10y = 800 meets coordinate axes.at A2<\/sub>(40, 0) and B2<\/sub>(0, 80).
\nThe region containing (0, 0) gives the solution set of inequation since (0, 0) satisfies the inequation.
\nx \u2265 0 ; y \u2265 0 represents the first quadrant, both lines intersects at P (30, 20).
\nThus the shaded area OA2<\/sub>PB1<\/sub> be the feasible region.
\nThe feasible region of the LPP is shaded in graph.
\n\"ML
\nThe coordinates of the vertices (Corner – points) of shaded feasible region OA2<\/sub>PB1<\/sub> are O (0, 0), A2<\/sub>(40, 0), P (30, 20) and B1<\/sub>(0, 50).
\nThe values of the objective of function at these points are given in the following table :<\/p>\n\n\n\n\n\n\n\n
Point (x2<\/sub>, y2<\/sub>)<\/td>\nValue of objective function
\nZ = 10,500x + 9,000y<\/td>\n<\/tr>\n
A2<\/sub> (40, 0)<\/td>\nZ = 10500 \u00d7 40 = 4,20,000<\/td>\n<\/tr>\n
P (30, 20)<\/td>\nZ = 10500 \u00d7 30 + 9000 \u00d7 20 = 4,95,000<\/td>\n<\/tr>\n
B1<\/sub> (0, 50)<\/td>\nZ = 10500 \u00d7 0 + 9000 \u00d7 50 = 4,50,000<\/td>\n<\/tr>\n
O (0, 0)<\/td>\nZ = 0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Here Zmax<\/sub> = 4,95,000 at x = 30 ; y = 20
\nThus, 30 hectares of land should be allocated to crop X and 20 hectares of land should be allocated to crop Y to maximize the profit.
\nThus, the required maximum profit that can be earned is \u20b9 4,95,000.
\nYes, I agree with the message that the protection of wildlife is utmost necessary to preserve the balance in environment.<\/p>\n

\"ML<\/p>\n

Question 16.
\nA manufacturer produces two products A and B. Both the products are processed on two different machines. The available capacity of the first machine is 12 hours and that of second machine is 9 hours. Each unit of product A requires 3 hours on both machines and each unit of product B requires 2 hours on first machine and 1 hour on the second machine. Each unit of product A is sold at a profit of \u20b9 7 and B at a profit of \u20b9 4, find the production level for maximum profit graphically. (Type CBSE Delhi 2016)
\nAnswer:
\nGiven data can be arranged in tabular form is as under :<\/p>\n\n\n\n\n\n\n
Product Machine<\/td>\nI<\/td>\nII<\/td>\nProfit (in \u20b9)<\/td>\n<\/tr>\n
A<\/td>\n3<\/td>\n3<\/td>\n7<\/td>\n<\/tr>\n
B<\/td>\n2<\/td>\n1<\/td>\n4<\/td>\n<\/tr>\n
Maximum availability capacity<\/td>\n12<\/td>\n9<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Let x units of product A and y units of product B are produced by manufacturer.
\nSince it is given that each unit of product A is sold at profit of \u20b9 5 and that of product B is sold at profit of \u20b9 6. Thus the total profit earned by manufacturer is selling x units of product A and y units of product B be 5x + 6y.
\nLet Z be the total profit \u2234 Z = 7x + 4y and our aim is to maximise Z. and corresponding constraints are : 3x + 2y \u2264 12 (Machine I constraint)
\n3x + y \u2264 9 (Machine II constraint)
\nx \u2265 0; y \u2265 0
\n[since no. of units of product A and B can\u2019t be negative]
\nFor region 3x + 2y \u2264 12 ; The line 3x + 2y = 12 meets coordinate axes at A (4, 0) and B (0, 6). Since (0,0) satisfies the given inequality. Thus region containing (0, 0) gives the solution set of given region.
\nFor region 3x + y \u2264 9 ; The line 3x + y = 9 meets coordinate axes at C (3, 0) and D (0, 9). Since (0, 0) lies on given inequality. Thus, the region containing (0, 0) gives the solution set of given region, x \u2265 0, y \u2265 0 represents the first quadrant of XOY plane.
\nBoth lines 3x + 2y = 12 and 3x + y = 9 intersects at P (2, 3). Here the shaded region be the feasible region OCPB and its corner points are O (0, 0) ; C (3, 0) ; P (2, 3) and B (0, 6). Now we evaluate Z at these comer points.<\/p>\n\n\n\n\n\n\n\n
Corner points<\/td>\nZ = 7x + 4y<\/td>\n<\/tr>\n
O (0, 0)<\/td>\n7 \u00d7 0 + 4 \u00d7 0 = 0<\/td>\n<\/tr>\n
C (3, 0)<\/td>\n7 \u00d7 3 + 4 \u00d7 0 = 21<\/td>\n<\/tr>\n
P (2, 3)<\/td>\n7 \u00d7 2 + 4 \u00d7 3 = 26<\/td>\n<\/tr>\n
B (0, 6)<\/td>\n7 \u00d7 0 + 4 \u00d7 6 = 24<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Zmax<\/sub> = 26 at B (0, 6) i.e. at x = 0 and y = 6
\nHence maximum profit of \u20b9 26 is earned when 0 units of product A and 6 units of product B are produced.
\n\"ML<\/p>\n

Question 16(Old).
\nOne kind of cake requires 200 g of flour and 25 g of fat, and another kind of cake requires vy00 g of flour and 50 g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no shortage of other ingredients used in making the cakes. Formulate the above as a linear programming problem and solve graphically.
\nAnswer:
\nLet x be the required number of one kind of cake and y be the required number of second kind of cakes that are made.
\n\"ML
\nLet Z be the total no. of cakes. Then Z = x + y
\nSince one unit of cake of one kind and second kind contain 200 g and 100 g of flour. Then x units of cake of first kind and y unit of cake of second and contain 200x and 100y of flour respectively but maximum flour available be (5 \u00d7 1000) then first constraint be, 2.00x + 100y < 5000
\n\u21d2 2x + y < 50
\nSince one unit of two types of cake contains 15 g and 50 g of fat required. Thus x and y units of both kind of cake contain, 25x and 50y gm of fat but maximum fat available be 1 kg. i.e. 25x + 50y < 1000 Then the mathematical model of the LPP is the follows : Maximize Z = x + y Subject to 200x + 100y \u2264 5000, 25x + 50y \u2264 1000 and x \u2265 0, y \u2265 0 To solve the LPP we convert the inequations into equations : 2x + y = 50 ; x + 2y = 40 ; x = 0 = y For region 2x + y \u2264 50 ; The line meet coordinate axes at A(25, 0) and E(0, 50). Since (0, 0) satisfies 2x + y \u2264 50. Thus the soln. set of 2x + y \u2264 50 containing (0, 0). For region x + 2y \u2264 40 ; The line x + 2y = 40 meet coordinate axes at D(40, 0) and C(0, 20). Since (0, 0) satisfies the inequation. Thus solution set of given inequation containing origin. Further x, y> 0 represents the first quadrant of xy region and both lines intersects at B(20, 10).
\nThe feasible region OABC of the LPP is shaded in graph.<\/p>\n

The coordinates of the vertices (Corner-points) of shaded feasible region OABC are given by 0(0, 0), A(25, 0), B(20, 10) and C(0, 20).
\nThe value of the objective of function at these points are given in the following table:<\/p>\n\n\n\n\n\n\n\n
Corner Point<\/td>\nValue of objective function Z = x + y<\/td>\n<\/tr>\n
A (25, 0)<\/td>\nZ = 25<\/td>\n<\/tr>\n
B(20, 10)<\/td>\nZ = 30<\/td>\n<\/tr>\n
C (0, 20)<\/td>\nZ = 20<\/td>\n<\/tr>\n
O (0, 0)<\/td>\nZ = 0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The maximum of 30 cakes can be made out of which 20 are of one kind and 10 are of IInd kind.<\/p>\n

Question 17.
\nA cottage industry manufactures pedastal lamps and wooden shades. Both products require machine time as well as craftsman time in making. The number of hour(s) for producing 1 unit of each and corresponding profit is given in the following table :
\nAnswer:
\nLet the manufacturer produces x pedestal lamps and y wooden shades to realise maximum profit. Since it is given that 1 unit of padestal lamp and wooden shade needs 1.5 hours and 3 hours machine time respectively. The factory has availability of not more than 42 hours of machine time. This can be represented by constraint 1.5x + 3y \u2264 42 => x + 2y \u2264 28 Similarly constraint associated with craftsman time be given as 3x + y \u2264 24
\nThus the given LPP can be represented mathematically as under: max Z = 30.x + 20y Subject to constraints; x + 2y \u2264 28 3x + y \u2264 24; x \u2265 0, y \u2265 0
\nFor region x + 2y \u2264 28 ; The line x + 2y = 28 meets x-axis at A (28, 0) and B (0,14). Clearly (0, 0) lies on given inequality. Thus region containing origin gives the solution set of given inequality.
\nFor region 3x + y \u2264 24 ; The line 3x + y = 24 meets coordinate axes at C(8, 0) and D (0, 24). Clearly (0, 0) lies on 3x + y \u2264 24. Thus region containing (0,0) gives the soln. set of given inequality.<\/p>\n

Now x \u2265 0, y \u2265 0 represents the positive quadrant of XOY plane.
\nBoth lines x + 2y = 28 and 3x + y = 24 intersects atP (4, 12)
\nClearly the bounded shaded region OCPBO represents the feasible region with comer points O (0, 0); C (8, 0); P (4, 12) and B (0, 14).
\n\"ML<\/p>\n\n\n\n\n\n\n\n
Corner points<\/td>\nZ = 30x + 20y<\/td>\n<\/tr>\n
O (0,0)<\/td>\nZ = o<\/td>\n<\/tr>\n
C (8, 0)<\/td>\nZ = 30 \u00d7 8 + 20 \u00d7 o = 240<\/td>\n<\/tr>\n
P (4,12)<\/td>\nZ = 30 \u00d7 4 + 20 \u00d7 12 = 120 + 240 = 360<\/td>\n<\/tr>\n
B (0, 14)<\/td>\nZ = 30 x 0 + 20 x 14 = 280<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Thus Z is maximum at x = 4 and y = 12
\nand Zmax<\/sub> = 360
\nHence the manufacturer produces 4 pedestal lamps and 12 wooden shades to realise max profit of \u20b9 360.<\/p>\n

Question 17(Old).
\nA company manufactures two types of toys A and B. A toy of type A requires 5 minutes for cutting and 10 minutes for assembling. A toy of type B requires 8 minutes for cutting and 8 minutes for assembling. There are 3 hours available for cutting and 4 hours available for assembling the toys in a day. The profit is \u20b9 50 each on a toy of Type A and \u20b9 60 each on a toy of type B. How many toys of each type should the company manufacture in a day to maximize the profit ? Use linear programming to find the solution. (ISC 2024)
\nAnswer:
\nLet number of toys of type A and B produced are x and y respectively to maximise the profit. Since, profits on each unit of toys A and B \u20b9 50 and \u20b9 60 respectively, so, profits on x units of toys A and y units of toy B are \u20b9 50x and \u20b9 60y respectively. Let Z be total profit.
\nThen Z = 50x + 60y
\nSince each unit of toy A and toy B require 5 min and 8 min for cutting. So x units of toy A and y units of toy B require 5x and 8y min, respectively but given maximum time available for cutting 3 x 60 = 180 min. so (first constraint) is given by
\n5x + 8y \u2264 180
\nSince each unit of toy A and toy B require 10 min and 8 min for assembling. So, x units of toy A and y units of toy B require lOx and 8y for assembling respectively but given maximum time available for assembling is 4 x 60 = 240 min, so (Second constraint) is given by
\n10x + 8y \u2264 240
\ni. e. 5x + 4y \u2264 120
\nHence the mathematical formulation of LPP is given below :
\nmaximize Z = 50x + 60y
\nSubject to constraints,
\n5x + 8y \u2264 180
\n[Since production can not be less than zero]
\nRegion 5x + 8y \u2264 180 : The line 5x + 8y = 180 meets coordinate axes at A1<\/sub>(36,0) and B1<\/sub>(0, \\(\\frac{45}{2}\\)) respectively.
\nRegion containing origin represents the solution set of 5x + 8y \u2264 180 as (0,0) satisfies 5x + 87 \u2264 180.
\n\"ML
\nRegion 5x + 4y \u2264 120 : The line 5x + 4y = 120 meets coordinate axes at C1<\/sub>(24, 0) and D1<\/sub>(0, 30) respectively.
\nRegion containing origin represents the solution set of 5x + 4y \u2264 120, as (0,0) satisfies 5x + 4y \u2264 120. Region x, y \u2265 0 : it represent first quadrant of xy plane.
\nThe shaded region OC1<\/sub>E1<\/sub>B1<\/sub> represents feasible region.
\nThe point E1<\/sub>( 12, 15) is the point of intersection obtained by solving 5x + 8y = 180 and 5x + 4y = 120 simultaneously.<\/p>\n\n\n\n\n\n\n\n
Corner Point<\/td>\nZ = 50x + 60y<\/td>\n<\/tr>\n
O(0, 0)<\/td>\n50(0) + 60(0) = 0<\/td>\n<\/tr>\n
C1(24, 0)<\/td>\n50(24) + 60(0) = 1200<\/td>\n<\/tr>\n
E1(12, 15)<\/td>\n50(12) + 60(15) = 1500<\/td>\n<\/tr>\n
B1(0, \\(\\frac{45}{2}\\))<\/td>\n50(0) + 60(\\(\\frac{45}{2}\\)) = 1350<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Clearly maximum Z = 1500 at x = 12, y = 15
\nThus required Number of toys A = 12, required no. of toys B = 15 & Maximum profit = \u20b9 1500<\/p>\n

Question 18.
\nAn aeroplane can carry a maximum of 200 passengers. Baggage allowed for a first class ticket is 30 kg and for an economy class ticket is 20 kg. Maximum capacity for the baggage is 4500 kg. The profit on each first class ticket is \u20b9 500 and on each economy class ticket is \u20b9 300. Determine how many tickets of each type must be sold to maximize the profit of the airline. Also find the maximum profit.
\nAnswer:
\nLet A andy be the number of first class and economy class tickets. Now it is given that, profit on each first class ticket is \u20b9 500 and on each economy class ticket is \u20b9 300.
\nThus, the profit gained by airline on selling A ticket of first class and y ticket of economy class be 500x + 300y.
\nLet Z be the total profit of airline.
\nThen Z = 500A + 300y and we want to maximize Z.
\nThe mathematical modelling of given LPP is given as follows : max Z = 500x + 300y
\nSubject to constriants ; x + y \u2264 200 [carry constraints]
\n30x + 20y \u2264 4500 [baggage constraint]
\ni. e. 3x + 2y \u2264 450
\nAlso, x \u2265 0; y \u2265 0 [Since first class and economy class tickets can\u2019t be negative]
\nFor region x + y \u2265 200 ; The line x + y = 200 meet coordinate axis at A (200, 0) and B(0, 200). Since (0, 0) lies on x + y \u2264 200.
\n\u2234 The region containing (0, 0) gives the soln. set of given inequation.
\nFor region 3x + 2y \u2264 450 ; The line 3x + 2y = 450 meet coordinate axis at C (150, 0) and D(0, 225). Since (0, 0) lies on given inequation. Thus region containing (0, 0) gives the soln. set of given inequation.
\nRegion x, y \u2265 0 represents the first quadrant of XOY plane.
\nBoth given lines intersects at P (50, 150).
\nThus the shaded region OCPB represents the feasible shaded region with corner points O (0, 0); C (150, 0) ; P (50, 150) and B (0,200).
\nWe evaluate Z at these corner points<\/p>\n\n\n\n\n\n\n\n
Corner point<\/td>\nZ = 500x + 300y<\/td>\n<\/tr>\n
O (0, 0)<\/td>\n0<\/td>\n<\/tr>\n
C (150, 0)<\/td>\n500 \u00d7 150 + 0 = 75000<\/td>\n<\/tr>\n
P (50, 150)<\/td>\n500 \u00d7 50 + 300 \u00d7 150 = 25000 + 45000 = 70000<\/td>\n<\/tr>\n
B (0, 200)<\/td>\n0 + 300 \u00d7 200 = 60, 000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"ML
\nZmax<\/sub> = 75,000 at C (150, 0) i.e. at x = 150 and y = 0
\nThus, required no. of first class ticket = 150
\nand required no. of economy class ticket = 0
\nand Maximum profit = \u20b9 75,000<\/p>\n

Question 18(Old).
\nA manufacturer manufactures two types of plastic bags A and B using two machines M1<\/sub> and M2<\/sub>. TO manufacture a bag of type A, machine M1<\/sub> is operated for 2 minutes and machine M2<\/sub> for 1 minute ; to manufacture a bag of type B, machine M1<\/sub> is operated for 3 minutes and machine M2<\/sub> for 4 minutes. Each machine can be operated for atmost 8 hours per day. A bag of type A is sold at a profit of \u20b9 0.50 and a bag of type B is sold at a profit of \u20b9 0.75. How many bags of each type should be manufactured and sold per day so as to maximise his profit. Form an L.P.P. and solve it graphically. Why plastic bags should not be used ? What values are being promoted ? (Value Based)
\nAnswer:
\nLet x be the required no. of plastic bags A and y be the required no. of plastic bag B be needed by the manufacturer to minimize his profit.
\nSince it is given that, bag A is sold at a profit of \u20b9 0.50 and bag B is sold at profit of \u20b9 0.75. So on selling x units of bag A gives profit 0.50x and y units of bag B gives profit 0.75y. Let Z be the total profit.
\nThen Z = 0.50x + 0.75y and we want to maximize Z.
\nTo manufacture bag A and B, machine M1<\/sub>, is operated for 2 min and 3 minutes and machine M1<\/sub>, can be operated for atmost 8 hours per day.
\n\u2234 The corresponding constraint is given by
\n2x + 3y < 8 x 60 = 480
\nTo manufacture bag A and B, machine M2<\/sub> is operated for 1 minute and 4 minute and also machine M2<\/sub> can be operated for atmost 8 hrs. per day.
\nSo, the corresponding constraint is given by x + 4y < 8 x 60 \u2014 480
\nThus, the mathematical modelling of given LPP is as under :
\nMax Z = 0.50+ 0.75y Subject to Constraints ;
\n2x + 3y \u2264 480
\nx + 4y \u2264 480
\nx, y \u2265 0 [Since no. of bags A and bag B cannot be negative]<\/p>\n

To solve LPP, we convert inequations into eqns :
\n2x + 3y = 480 ;
\nx + 4y = 480 ;
\nx = 0 = y
\nFor region 2x + 3y \u2264 480 ;
\nThe line 2x + 3y = 480 meets coordinate axis at A (240, 0) and B (0, 160).
\nSince (0, 0) lies on 2x + 3y \u2264 480. So region containing origin gives the solution set of given inequation.<\/p>\n

For region x + 4y \u2264 480 :
\nThe line x + 4y = 480 meets coordinate axis at C(480, 0) and D(0, 120).
\nSince (0, 0) lies on x + 4y \u2264 480. Thus region containing (0, 0) gives the soln. set of given inequation.
\nx \u2265 0, y \u2265 0 represents the first quadrant of XOY plane. Both lines intersects at P (96, 96).
\nThus the shaded region OAPD gives the bounded feasible region with corner points O (0, 0), A (240, 0) ; P (96, 96) and D (0, 120).
\n\"ML<\/p>\n\n\n\n\n\n\n\n
Corner Point<\/td>\nZ = 0.5x + 0.75y<\/td>\n<\/tr>\n
O (0, 0)<\/td>\n0<\/td>\n<\/tr>\n
A1<\/sub>(240, 0)<\/td>\n0.5 \u00d7 240 =120<\/td>\n<\/tr>\n
P(96, 96)<\/td>\n0.5 \u00d7 96 + 0.75 \u00d7 96 =120<\/td>\n<\/tr>\n
D (0,120)<\/td>\n0.75 \u00d7 120 = 90<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Clearly Zmax<\/sub> =120 at x = 240 or y = 0 or at x = y = 96.
\nRequired no. of plastic bag A = 96 = Required no. of plastic bag B = 96 and Max profit = 120 ; Since plastic bags are biodegradable so they are harmful for environment. So these bags are also block drainage system. Hence values promoted are love for environment and increase the productivity of soil.<\/p>\n

Question 19.
\nA carpenter has 90,80 and 50 running feet respectively of teak wood, plywood and rosewood which is used to produce product A and product B. Each unit of product A requires 2, 1 and 1 running feet and each unit of product B requires 1, 2 and 1 running feet of teak wood, plywood and rosewood respectively. If product A is sold for Rs. 48 per unit and product B is sold for Rs. 40 per unit, how many units of product A and product B should be produced and sold by the carpenter, in order to obtain the maximum revenue. Formulate the above as a linear programming problem and solve it, indicating clearly the feasible region in the graph. (ISC 2019)
\nAnswer:
\nLet x units of product A and y units of product B should be produced and sold by the carpenter in order to obtain maximum revenue.
\nIt is given that product A is sold for Rs. 48 per unit and product B is sold for Rs. 40 per unit.
\nLet Z be the maximum profit of carpenter Then Z = 48x + 40y
\nThe given data of given LPP is tabulated as under :
\n\"ML
\nThe mathematical modeling of given LPP is given as under Max Z = 48x + 40y
\nSubject to constraints :
\n2x + y \u2264 90 ; x + 2y \u2264 80; x + y \u2264 50
\nx \u2265 0 y \u2265 0 [since no. of units can\u2019t be negative]
\nFor region 2x + y = 90; the line 2x + y = 90 meets coordinate axes at A (45, 0) and B(0, 90). Clearly (0, 0) lies on given inequality. Thus region containing (0, 0) gives the solution set of given inequality.
\nFor region x + 2y \u2264 80; The line x + 2y = 80 meets coordinate axes at C (80, 0) and D(0, 40). Clearly (0, 0) satisfies the inequality. Thus, region containing (0, 0) gives the solution set of given inequality.
\nFor region x + y \u2264 50 ; The line x + y = 50 intersects coordinate axes at E(50, 0) and F(0, 50). Clearly (0, 0) satisfies the given inequality. Thus region containing (0, 0) gives the solution set of given inequality.
\nRegion x \u2265 0, y \u2265 0 represents the first quadrant of XOY plane
\nThe lines 2x + y = 90 and x + 2y = 30 intersects at P(\\(\\left(\\frac{100}{3}, \\frac{70}{3}\\right)\\))
\nThe lines x + 2y = 80 and x + y = 50 intersects at Q(20, 30).
\nThe lines 2x + y = 90 and x + y = 50 intersects at R(40, 10).
\nClearly the bounded shaded region represents the feasible region OARQDO and its comer points are 0(0, 0); A(45, 0); R(40, 10); Q(20, 30); and D(0, 40). We evaluate Z at these comer points and tabulated given below :
\n\"ML<\/p>\n\n\n\n\n\n\n\n\n
Corner points<\/td>\nZ = 48x + 40y<\/td>\n<\/tr>\n
0(0, 0)<\/td>\n0<\/td>\n<\/tr>\n
A(45, 0)<\/td>\n48 \u00d7 45 = 2160<\/td>\n<\/tr>\n
R(40,10)<\/td>\n48 \u00d7 40 + 40 \u00d7 10 = 2320 4<\/td>\n<\/tr>\n
Q(20, 30)<\/td>\n8 \u00d7 20 + 40 \u00d7 30 = 2160<\/td>\n<\/tr>\n
D(0, 40)<\/td>\n= 1600<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Clearly Zmax<\/sub> = 2320 at x = 40 and y = 10.
\nThus 40 units of produce A and 10 units of product B are to be produce and sold by carpenter to realize maximum profit Rs. 2320.<\/p>\n

\"ML<\/p>\n

Question 20.
\nA diet is to contain atleast 80 units of Vitamin A and 100 units of minerals. Two foods F1<\/sub> and F2<\/sub> are available costing \u20b9 5 per unit and \u20b9 6 per unit respectively. One unit of food F1<\/sub> contains 4 unitsof vitamin A and 3 units of minerals whereas one unit of food F2<\/sub> contains 3 units of vitamin A and 6 units of minerals. Formulate this as a linear programming problem. Find the minimum cost of diet that consists of mixture of these two foods and also meets minimum nutritional requirement.
\nAnswer:
\nLet x units of food F1<\/sub> and y units of food F2<\/sub> respectively used to minimise the cost. Given cost of 1 unit of food F1<\/sub> and food F2<\/sub> be \u20b9 5 and \u20b9 6 respectively. Then cost of x units of food F1<\/sub> and y units of food F2<\/sub> be \u20b9 5x and \u20b9 6y respectively. Let Z be the total cost of diet.
\nThen Z = 5x + 6y
\nThe given problem can be summarized in tabular form is as under :
\n\"ML
\nThus the mathematical modelling of given LPP is given as under :
\nMin Z = 5x + 6y
\nSubject to Constraints :
\n4x + 3y \u2265 80
\n3 x + 6y \u2265 100
\nx \u2265 0, y \u2265 0
\nFor region 4x + 3y \u2265 80 ; The line 4x + 3y = 80 meets x-axis at A (20,0) andy-axis at B(0, \\(\\frac{80}{3}\\)) . Clearly (0, 0) does not satisfies 4x + 3y \u2265 80.<\/p>\n

Then Z = 50x + 70y
\nThe given problem can be summarised in tabular form is as under :
\n\"ML
\n\"ML
\nThus, the mathematical modelling of given LPP is given as under :
\nMin. Z = 50x + 70y Subject to constraints ;
\n2x + y \u2265 8
\nx + 2y \u2265 0
\nx \u2265 0, y \u2265 0
\nFor region 2x + y \u2265 8;
\nThe line 2x + y = 8 meets coordinate axes at A (4,0) and B (0, 8). Clearly (0, 0) does not satisfies 2x + y > 8. Thus region not containing origin gives the solution set of 2x + y > 8.
\nFor region x + 2y > 10 ; The line x + 2y = 10 meets coordinate axes at C (10, 0) and D(0, 5).
\nClearly (0,0) does not satisfies x + 2y > 10. Thus region not containing (0, 0) gives the soln. set of given inequality.
\nFurther x \u2265 0, y > 0 represents the first quadrant of XOY plane.
\nThe lines 2x + y = 8 and x + 2y = 10 intersects at P (2, 4).
\nThe unbounded shaded region BPC represents the feasible region with comer points B (0, 8); P (2, 4) and C (10, 0).<\/p>\n\n\n\n\n\n\n
Corner points<\/td>\nZ = 50x + 70y<\/td>\n<\/tr>\n
B (0, 8)<\/td>\nZ = 50 \u00d7 0 + 70 \u00d7 8 = 560<\/td>\n<\/tr>\n
P (2, 4)<\/td>\nZ = 50 \u00d7 2 + 70 \u00d7 4 = 380<\/td>\n<\/tr>\n
C (10, 0)<\/td>\nZ = 50 \u00d7 10 + 70 \u00d7 0 = 500<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Here smallest value of Z is 380 since the feasible region is unbounded. So we have to check whether Z is minimum or not. For this, we draw the line 50x + 70y = 380 and check open half plane 50x + 70y < 380 have common points in feasible region. Clearly the open half plane 50x + 70y < 380 has no common points in feasible region so the smallest value Z is minimum.
\n\u2234 Zmin<\/sub> = 380 at x = 2 andy = 4
\nThus 2 kg of food I and 4 kg of food II are mixed to realize minimum cost.<\/p>\n

Question 22.
\nReshma wishes to mix two types of foods P and Q in such a way that the vitamin contents of the mixture contains atleast 8 units of vitamin A and 11 units of vitamin B. Food P costs ? 60\/kg and food Q costs ? 80\/kg. Food P contains 3 units\/kg of vitamin A and 5 units\/kg of vitamin B while food Q contains 4 units\/kg of vitamin A and 2 units\/kg of vitamin B. Determine the minimum cost of the mixture. (NCERT)
\nAnswer:
\nLet x kg of food P and y kg of food Q are mixed together to make the mixture and are required to minimise the cost of the mixture.
\nThen the required mathematical model of the LPP is given as follows :
\nMinimize Z = 60x + 80y Subject to contraints,
\n3x + 4y \u2265 8,
\n5x + 2y \u2265 11
\nand x \u2265 0, y \u2265 0
\nTo solve the LPP we convert the inequations into equations as follows :
\n3x + 4y = 8
\n5x + 2y= 11;
\nx = 0 = y
\nFor region 3x + 4y \u2265 8; The line 3x + 4y = 8 meets coordinate axes at A(8\/3, 0) and D(0, 2). On joining these two points to get the line. Since (0,0) do not satisfies the inequation. Hence the solution set of inequation does not contain (0,0).
\n\"ML
\nFor region 5x + 2y \u2265 11: The line 5x + 2y= 11 meet coordinate axes at E (\\(\\frac{11}{5}\\), 0) and C(0, \\(\\frac{11}{2}\\)) Since (0,0) satisfies the inequation and solution set of inequation not containing the origin.
\nAlso x \u2265 0, y \u2265 0 represent the first quatrant.
\nBoth lines intersects at B(2, \\(\\frac{1}{2}\\))
\nThe feasible region ABC of the LPP is shown by shaded area in graph. The coordinates of the corner-points of shaded feasible region ABC are
\nThe values of the objective function at these corner points are given in the following table :<\/p>\n\n\n\n\n\n\n
Corner Point<\/td>\nValue of objective function
\nZ = 60x + 80y<\/td>\n<\/tr>\n
A(\\(\\frac{8}{3}\\), 0)<\/td>\nZ = 160<\/td>\n<\/tr>\n
B(2, \\(\\frac{1}{2}\\))<\/td>\nZ = 160<\/td>\n<\/tr>\n
C(0, \\(\\frac{11}{2}\\))<\/td>\nZ = 440<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The minimum value ofthe mixture is \u20b9 160 at all points on the line segment joining points (\\(\\frac{8}{3}\\), 0) and (2, \\(\\frac{1}{2}\\))<\/p>\n

Question 22(Old).
\nA cottage industry manufactures pedestal lamps and wooden shades, each requiring the ^ use of a grinding\/cutting machine and a sprayer. It takes 2 hours on the grinding\/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp while it takes 1 hour on the grinding\/cutting machine and 2 hours on the sprayer to manufacture a shade.
\nTo keep the noise and dust pollution under prescribed limits, a sprayer can be used for a maximum of 20 hours per day while cutting and grinding machine can be used for a maximum of 12 hours per day. The profit from the sale of a lamp is ? 5 and that from sale of a shade is ? 3. Assuming that manufacturer can sell all the pedestal lamps and shades that are made, how should he schedule his daily production to maximise profit ? Make it as an L.P.P. and solve it graphically. Which value is indicated in the above question ?
\nAnswer:
\nLet the cottage industry manufacture x pedestal lamps and y wooden shades.
\nTherefore, x \u2265 0 and y \u2265 0
\nThe given infromation can be tabulated as follows :
\n\"ML
\nThe profit on a lamp is ? 5 and on the shades is ? 3. Therefore, the constraints are
\n2x + y \u2264 12
\n3x + 2y\u2264 20
\nTotal profit, Z = 5x + 3y, we have to maximise Z.
\nThe mathematical formulation of the given problem is given below :
\nMaximize Z = 5x + 3y
\nsubject to the constraints,
\n2x + y \u2264 12
\n3x + 2y \u2264 20
\nx, y \u2265 0
\nRegion 2x + y \u2264 12 ; The line 2x+y= 12 meet coordinate axes at A(6,0) and B1<\/sub>(0,12). region containing (0,0) gives the solution set of 2x + y \u2264 12 since (0,0) lies on 2x + y \u2264 12.
\nRegion 3x + 2y \u2264 20 : The line 3x + 2y = 20 meet coordinate axes at A2<\/sub>(\\(\\frac{20}{3}\\),0) and C(0, 10). The region containing (0,0) be the solution set of given inequation.
\nThe region x \u2265 0, y \u2265 0 represents the first quadrant of xy plane.
\nBoth lines intersects at B(4, 4).
\nThe shaded area gives feasible region OABC determined by the system of constraints is as follows. The corner points are 0(0,0), A(6,0), B(4,4) and C(0,10).
\nThe value of Z at these corner points are as follows:
\n\"ML<\/p>\n\n\n\n\n\n\n\n
Corner Point<\/td>\nZ = 5x + 3y<\/td>\n<\/tr>\n
O(0, 0)<\/td>\n0<\/td>\n<\/tr>\n
A1<\/sub>(6, 0)<\/td>\n30<\/td>\n<\/tr>\n
P(4, 4)<\/td>\n32 \u2192 Maximum<\/td>\n<\/tr>\n
B2<\/sub>(0, 10)<\/td>\n30<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The maximum value of Z is 32 at (4,4).
\nThus, the manufacturer should produce 4 pedestal lamps and 4 wooden shades to maximize his profits i.e. noise and dust pollution be always under prescribed limits.<\/p>\n

Question 23.
\nA farmer has a supply of chemical fertiliser of type A which contains 10% nitrogen and 6% phosphoric acid and B consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 7 kg of nitrogen and 7 kg of phosphoric acid for her crop. If A costs \u20b9 5\/kg and B costs \u20b9 8\/kg, determine how much of each type of fertiliser should be used so that neutrient requirements are met at a minimum cost. What is the minimum cost ? Find the feasible region in the graph (NCERT)
\nAnswer:
\nLet x kg of fertiliser A and y kg of fertiliser B should be required to minimise the cost.
\nGiven information can be put in tabular form as below :<\/p>\n\n\n\n\n\n\n
Fertilizers<\/td>\nA<\/td>\nB<\/td>\nMin. Requirements<\/td>\n<\/tr>\n
Nitrogen<\/td>\n0.1x<\/td>\n0.05y<\/td>\n7 kg<\/td>\n<\/tr>\n
Phosphoric<\/td>\n0.06x<\/td>\n0.1y<\/td>\n7 kg<\/td>\n<\/tr>\n
Min. Cost<\/td>\n5x<\/td>\n8y<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Let Z be the total cost \u2234 Z = 5x + 8y
\nSo our aim is to minimize Z
\nThen the mathematical modelling of the LPP is given as follows :
\nMinimize Z = 5x + 8y
\nSubject to constraints;
\n0.1x + 0.05y \u2265 7
\n0.06x + 0.1y \u2265 7
\nand x, y \u2265 0
\nFor region 0.1x + 0.05y \u2265 7 ;
\nThe line 0.1x + 0.05y = 7 meet coordinate axes at A (70, 0) and B(0, 140). Since (0, 0) does not lies on given region.
\nThus the region not containing (0, 0) gives the solution set of given region.
\nRegion 0.06x + 0.01y \u2265 7 ;
\nThe line 0.06x + 0.1y = 7 meet coordinate axes at C(\\(\\frac{350}{3}\\), 0) and
\nD(0, 70). Since (0, 0) does not satisfies the set of given region.
\nThus the region not containing (0, 0) gives the solution set of given region.
\nFurther x \u2265 0, y \u2265 0 represents the first quadrant of XOY plane.
\n\"ML
\nFurther both lines 0.1x + 0.05y = 7 and 0.06x + 0.1y = 7 intersects at point P(50, 40). Clearly the shaded area CPB be the feasible region and feasible region in unbounded.<\/p>\n\n\n\n\n\n\n
Corner Point<\/td>\nZ = 5x + 8y<\/td>\n<\/tr>\n
C(\\(\\frac{350}{3}\\), 0)<\/td>\nZ = 5 \u00d7 \\(\\frac{350}{3}\\) + 8 \u00d7 0 = \\(\\frac{1750}{3}\\)<\/td>\n<\/tr>\n
P(50, 40)<\/td>\nZ = 5 \u00d7 50 + 8 \u00d7 40 = 570<\/td>\n<\/tr>\n
B(0, 140)<\/td>\nZ = 5 \u00d7 0 + 8 \u00d7 140 = 1120<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Here smallest value of Z be 570 occurs atx = 50 andy = 40. Since the feasible region is unbounded. So we check smallest value of Z is minimum or not. For this, we draw the line 5x + 8y = 570 and check whether open half plane 5x + 8y < 570 have common points in feasible region or not. Since there are no common points in feasible region.
\nThus smallest value be the minimum value.
\nZmin<\/sub> = 500 at x = 50 and y = 40
\nHence 50 kg of fertiliser A and 40 kg of fertiliser B should be bought to get the minimum cost of \u20b9 570.<\/p>\n

Question 24.
\nA toy company manufactures two types of dolls, A and B. Market tests and available resources indicated that the combined production level should not exceed 1200 dolls per week and the demand of the dolls of B type is atmost half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by atmost 600. If the company makes profit of \u20b9 12 and \u20b9 16 per doll respectively on dolls A and B, how many of each type should be produced weekly in order to maximize the profit ? (NCERT)
\nAnswer:
\nLet x be the required number of dolls of type A and y be the required number of dolls of type B should be produced to earn the maximum profit.
\nThen the mathematical modelling of the LPP is given as follows :
\nMaximize Z = 12x+ 16y
\nSubject to constraints ;
\nx + y \u2264 1200
\n\\(\\frac{1}{2}\\)x – y \u2265 0
\nx – 3y \u2264 600
\nx \u2265 0, y \u2265 0
\nTo solve the LPP we convert the given inequations into equations x + y = 1200
\nx + y = 1200
\n\\(\\frac{1}{2}\\)x – y = 0
\nx – 3y = 600
\nRegion x + y \u2264 1200 ; The lin ex+y= 1200 meets coordinate axes at A1<\/sub> (1200, 0) and B1<\/sub>(0,1200). Since (0, 0) satisfies the x + y \u2264 1200. So the region containing (0, 0) gives the solution set of given inequation.
\nRegion \\(\\frac{1}{2}\\)x – y > 0; The line \\(\\frac{x}{2}\\) – y = 0 passes through origin.
\nRegion x – 3y \u2264 600 : The line x -3y = 600 meets the coordinate axes at A2<\/sub> (600,0) and B2<\/sub> (0, – 200). Since (0, 0) satisfies the inequations.
\nHence the region containing (0, 0) gives the solution set of inequation.
\nAlso, x \u2265 0, y \u2265 0 represent the first quadrant of xoy plane.
\nThe lines x + y = 1200 and \\(\\frac{x}{2}\\) – y = 0 intersect at P (800, 400).
\nThe lines x + y = 1200 and x – 3y = 600 intersect at Q (1050, 150)
\nThe feasible region of the LPP is shaded in graph.
\nThe shaded area OA2QP gives the feasible region
\nThus, the coordinates of the vertices of (Comer – points) of shaded feasible region OA2<\/sub>PQ are O (0, 0) ; A2<\/sub> (600, 0), Q (1050, 145) and P (800, 400).
\n\"ML
\nThe values of the objective function at corner points are given in the following table:<\/p>\n\n\n\n\n\n\n\n
Corner Point (x, y)<\/td>\nValue of objective function Z = 12x + 16y<\/td>\n<\/tr>\n
A2<\/sub> (600, 0)<\/td>\nZ = 12 \u00d7 600 + 16 \u00d7 0 = 7200<\/td>\n<\/tr>\n
Q (1050, 145)<\/td>\nZ= 12 \u00d7 1050 + 16 \u00d7 145 = 14920<\/td>\n<\/tr>\n
P(800, 400)<\/td>\nZ = 12 \u00d7 800 + 16 \u00d7 400 = 16000<\/td>\n<\/tr>\n
O (0, 0)<\/td>\nZ = 0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Clearly Zmax<\/sub> = 16000 at x = 800 and y = 400
\nThus, the toy company should manufacture 800 dolls of type A and 400 tolls of type B to earn maximum profit.
\nThus, the required maximum profit that can be earned by company is \u20b9 16,000.<\/p>\n

Question 24(Old).
\nA small scale factory makes two types of dolls. One doll of type I takes T5 hours of electronic machine and 3 hours of hand operated machine ; one doll of type II takes 3 hours of electronic machine and 1 hour of hand operated machine. In a day, the factory has the ability of atmost 42 hours of electronic machines and 24 hours of hand operated machines. If the profit on one doll of type I is \u20b9 20 and on one doll of type II is \u20b9 30, find the number of dolls of each type that the factory should manufacture to earn maximum profit. Make it as an L.P.P. and solve graphically. Why are small scale industries important in India ? What values are being promoted by establishing small scale industry ? (Value Based)
\nAnswer:
\nLet x be the no. of dolls of type I and y be the no. of dolls of type II that a factory made and sells. The profit on selling of doll I and II are \u20b9 20 and \u20b9 30 respectively. So profit gained by comnpany on selling x units of doll I and y units of doll II be 20x + 30y.
\nLet Z be the total profit of company. \u2234 Z = 20x + 30y we have to maximize Z.
\nSince it is given that, one unit of doll I and II takes 1 -5 hours and 3 hours of electronic machine and factory has the ability of atmost 42 hrs of electronic machine. Thus first constraint be ;
\n1.5x + 3y \u2264 42 => x + 2y \u2264 28
\nAlso it is givne that, one unit of doll I and II takes 3 hours and 1 hour of hand operated machine. Also the factory has the ability of atmost 24 hrs. of hand operated machines
\n\u2234 3x + y \u2264 24
\nThe mathematical formulation of given LPP is as under ;
\nMax Z = 20x + 30y
\nSubject to constraints ;
\nx + 2y \u2264 28
\n3x + y \u2264 24
\nx, y \u2265 0 [since, dolls of I and II can\u2019t be negative]
\nTo solve LPP, we convert inequations into eqn\u2019s :
\nFor region x + 2y \u2264 28 ; The linex + 2y = 28 meets coordinate axis at A (28,0) and B(0, 14). Since (0, 0) lies on x + 2y \u2264 28. So the region containing (0, 0) gives the soln. set of given inequation. For region 3x + y \u2264 24 ; The line 3x+y = 24 meets coordinate axis at C (8, 0) and D (0, 24). Since (0, 0) lies on 3x + y \u2264 24.
\nSo region containing (0, 0) gives the soln. set of 3x + y \u2264 24.
\nRegion x \u2265 0, y \u2265 0 represents the first quadrant of XOY plane.
\nBoth lines intersects at P (4, 12).
\nThus, the shaded region OCPD gives the feasible and bounded region with comer points 0(0, 0); C(8, 0); P(4, 12) and D(0, 14).
\n\"ML
\nWe evaluate Z at these corner points<\/p>\n\n\n\n\n\n\n\n
Corner point<\/td>\nZ = 20x + 30y<\/td>\n<\/tr>\n
O (0, 0)<\/td>\n0<\/td>\n<\/tr>\n
C (8, 0)<\/td>\n160<\/td>\n<\/tr>\n
P(4, 12)<\/td>\n80 + 360<\/td>\n<\/tr>\n
B (0, 14)<\/td>\n0 + 30 \u00d7 14 =\u00a0 420<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Zmax<\/sub> = 440 at P (4, 12) i.e. at A = 4 and y = 12
\n\u2234 required no. of doll of type 1 = 4; required no. of doll II = 12 and Maximum profit = \u20b9 440.
\nSince small scale Industry needs low investment and generate more employment and hence progression of the country. Thus values promoted are employment generation and helping in removing poverty.<\/p>\n

Question 25.
\nThe feasible region for an L.P.P. is shown in the adjoining figure.
\n\"ML
\n(i) Write the constraints for the L.P.P.
\n(ii) Write the coordinates of the points A, B and C.
\n(iii) Find the maximum value of the objective function Z = 30x + 50y.
\nAnswer:
\n(i) The eqn. of AD is
\n\\(\\frac{x}{24}+\\frac{y}{24}\\) = 1 i.e. x + y = 24
\nThe eqn. of EC is
\n\\(\\frac{x}{32}+\\frac{y}{16}\\) = 1 \u21d2 x + 2y = 32
\nAs, (0, 0) lies in x + y \u2264 24 and (0, 0) also satisfies x + 2y \u2264 32.
\nThus associated constraints for given L.P.P. are ; x + y \u2264 20 ; x + 2y \u2264 32, x \u2265 0, y \u2265 0
\n(ii) Clearly the coordinates of A are (24, 0) and that of C are (0, 16). The lines x + y = 24 and x + 2y = 32 intersects at B (16, 8).
\n(iii) The shaded bounded region represents the feasible region with comer points O (0,0); A (24,0); B (16, 8) and C (0, 16).<\/p>\n\n\n\n\n\n\n\n
Corner points<\/td>\nZ = 30x + 50y<\/td>\n<\/tr>\n
0 (0, 0)<\/td>\nZ = 30 \u00d7 0 + 50 \u00d7 0 = 0<\/td>\n<\/tr>\n
A (24,0)<\/td>\nZ = 30 \u00d7 24 + 50 \u00d7 0 = 720<\/td>\n<\/tr>\n
B (16, 8)<\/td>\nZ = 30 \u00d7 16 + 50 \u00d7 8 = 480 + 400 = 880<\/td>\n<\/tr>\n
C (0, 16)<\/td>\nZ = 30 \u00d7 0 + 50 \u00d7 16 = 800<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Clearly Zmax<\/sub> = 880 at B (16, 8)
\ni.e. at x = 16 and y = 8<\/p>\n

\"ML<\/p>\n

Question 26.
\nThe feasible region for an L.P.P. is shown in the adjoining figure. The line CB is parallel to OA. Answer the following :
\n\"ML
\n(i) Write the constraints for the L.P.P.
\n(ii) Write coordinates of the point B.
\n(iii) Find the minimum value of the objective function Z = 3x-4y
\nAnswer:
\n(i) The eqn. of line CB be given by
\ny – 4 = \\(\\frac{16-4}{6-0}\\)(x – o)
\n\u21d2 y – 4 = 2x \u21d2 y = 2x + 4
\nThe eqn. of line OA be given by 12-0
\ny – 0 = \\(\\frac{12-0}{6-0}\\)(x – 0)
\n\u21d2 y = 2x
\nClearly (0, 0) satisfies y – 2x \u2264 4 and (0, 0) satisfies y \u2265 2x.
\nAlso (0, 0) satisfies x \u2264 6.
\nThus the associated constraints for given L.P.P is as under : y – 2x \u2264 4; y \u2265 2x, x \u2264 6; x \u2265 0, y \u2265 0<\/p>\n

(ii) Point B be the point of intersection of lines x = 6 and y – 2x = 4 ,Coordinates of B are (6, 16).
\n(iii) The bounded shaded region OCBAO represents the feasible region with comer points O (0,0); C (0, 4); B (6, 16) and A (6, 12).<\/p>\n\n\n\n\n\n\n\n
Corner points<\/td>\nZ = 3x \u2013 4y<\/td>\n<\/tr>\n
0 (0, 0)<\/td>\nZ = 3 \u00d7 0 – 4 \u00d7 0 = 0<\/td>\n<\/tr>\n
C (0, 4)<\/td>\nZ = 3 \u00d7 0 – 4 \u00d7 4 = – 16<\/td>\n<\/tr>\n
B (6, 16)<\/td>\nZ = 3 \u00d7 6 – 4 \u00d7 16 = – 46<\/td>\n<\/tr>\n
A (6,12)<\/td>\nZ = 3 \u00d7 6 – 4 \u00d7 12 = -30<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Question 26(Old).
\nTwo tailors P and Q earn \u20b9 150 and \u20b9 200 per day respectively. P can stich 6 shirts and 4 trousers a day, while Q can stich 10 shirts and 4 trousers per day. How many days should each work to produce atleast 60 shirts and 32 trousers at minimum labour cost? (ISC 2012) Sol. {suppose tailor A and B work for x and y days respectively to minimise the cost.
\nAnswer:
\nSuppose tailor A and B work for x and y days respectively to minimise the cost.
\nSince tall or A and B earn \u20b9 150 and \u20b9 200 respectively So, tailor A and B earning for x and y days be \u20b9 150 x and 200y respectively, let Z denote maximum profit that gives minimum labour cost Then Z = 150x + 200y
\nSince, Tailor A and B stitch 6 and 10 shirts respectively in a day, Thus tailor A can stitch 6x and B can stitch 10y shirts in x andy days respectively, but it is desired to produce 60 shirts at least. Then (first constraint) is given by
\n6x + 10y \u2265 60
\n3x + 5y \u2265 30
\nSince, Tailor A and B stitch 4 pants per day each, so tailor A can stitch 4x and B can stitch 4y pants in x andy days, respectively, but it is desired to produce at least 32 pants, Then (second constraint) is given by
\n4x + 4y \u2265 32
\nx + y \u2265 8
\nHence, The required mathematical formulation of LPP is, as given below Min. Z= 150.x+ 200y
\nSubject to constraints,
\n3x + 5y \u2265 30
\nx + y \u2265 8
\nx, y \u2265 0 [Since x and y not be less than zero]
\nRegion 3x + 5y \u2265 30: line 3x + 5y = 30 meets coordinate axes at A1<\/sub>(10,0) and B2<\/sub>(0,6) respectively. Region not containing origin represents the solution set of inequation 3x + 5y > 30, as (0,0) does not satisfy 3x + 5y \u2265 30.
\nRegion x + y \u2265 8: line x + y = 8 meets axes at C1<\/sub>(8,0) and D1<\/sub>(0,8) respectively. Region not containing origin represents the solution set of inequation x + y \u2265 8 as (0,0) does not satisfy x + y \u2265 8.
\n\"ML
\nRegion x, y \u2265 0 : it represent the first quadrant in the xy plane,
\nboth given lines 3x + 5y = 30 and x + y = 8 intersects at E1<\/sub>(5, 3).
\nUnbounded shaded region A1<\/sub>E1<\/sub>D1<\/sub> represents the feasible region with corner points A1<\/sub>(10,0), E1<\/sub>(5, 3) and D1<\/sub>(0.8).<\/p>\n\n\n\n\n\n\n
Corner point<\/td>\nZ = 150x + 200y<\/td>\n<\/tr>\n
A1<\/sub>(0, 0)<\/td>\n150(10) + 200(0) = 1500<\/td>\n<\/tr>\n
E1<\/sub> (5, 3)<\/td>\n150(5) + 200(3) = 1350<\/td>\n<\/tr>\n
D1<\/sub>(0, 8)<\/td>\n150(0) + 200(8) = 1600<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Smallest value of Z is 135, Since the feasible region is unbounded. We have to deck whether the value of Z i.e. 135 is minimum or not. For this we draw a line 150x + 200y = 1350 and check whether the open half plane 150x + 200y < 1350 have common points in feasible region or not. Now open half plane 150x + 200y < 1350 has no point in common with feasible region, so smallest value is the minimum value. So,
\nZmin = 1350, at x = 5, y = 3
\nThus, Tailor A should work for 5 days and B should work for 3 days.<\/p>\n

Question 27.
\nAn oil company has two depots A and B with capacities of 7000 litre and 4000 litre respectively. The company is to supply oil to three petrol pumps D, E and F, whose ,y requirements are 4500 litre, 3000 litre and 3500 litre respectively. The distance (in km) between the depots and the petrol pumps is given in the following table :
\n\"ML
\nAssuming that the transportation cost per km is \u20b9 1 per litre, how should the delivery be scheduled in order that the transportation cost is minimum ? What is the minimum cost? (NCERT)
\nAnswer:
\nLet x and y litres of oil be supplied from A to the petrol pumps, D and E. Then (7000-x-y) litres will be supplied from A to petrol pump F.
\nSince it is given that, the requirement at petrol pump D is 4500 L. Since x L are transported from depot A, the remaining (4500 -x) L will be transported from petrol pump B.
\nSimilarly, (3000 -y) Land 3500 -(7000- x-y) = (x +y- 3500) L will be transported from depot B to petrol pump E and F respectively.
\nThe given problem can be represented diagramatically given as follows.
\n\"ML
\nx \u2265 0, y \u2265 0 and (7000 – x – y) \u2265 0 x \u2265 0, y \u2265 0 and x + y \u2264 7000 i.e. 4500 – x \u2265 0, 3000 – y \u2265 0, and x + y – 3500 \u2265 0 Thus, x < 4500, y \u2264 3000, and x + y \u2265 3500
\nGiven cost of transporting 10 L of petrol = Re 1
\nCost of transporting 1 L of petrol = \u20b9 \\(\\frac{1}{10}\\)
\nTherefore, total transportation cost is given by,
\nZ = \\(\\frac{7}{10}\\) x x + \\(\\frac{6}{10}\\) y + \\(\\frac{3}{10}\\)(7000 – x – y) + \\(\\frac{4}{10}\\)(4500-x) + \\(\\frac{4}{10}\\)(3000 – y) + \\(\\frac{2}{10}\\)(x + y – 3500)
\ni.e. Z = 30x + 0.1y + 3950
\nThe given problem can be formulated as follows :
\nMinimize Z = 0.3x + 0.1y + 3950
\nsubject to the constraints,
\nx +y \u2264 7000
\nx \u2264 4500
\ny \u2264 3000
\nx + y \u2265 3500
\nx, y \u2265 0
\nThe feasible region determined by the constraints is as follows.
\nFor region x + y \u2264 7000 ; The line x + y = 7000 meets coordinate axes at A1<\/sub>(7000, 0) and B1<\/sub>(0, 7000). Region containing (0,0) represents the solution set of inequation.
\nFor region x \u2264 4500; The line x = 4500 is parallel to y-axis meeting x-axis at C,(4500,0).
\nSince (0. 0) lies on x \u2264 4500.
\nThus region containing (0,0) represents the solution set of inequation x \u2264 4500.
\nForregiony \u2264 3000 ; The line y \u2264 3000 be a line parallel to x-axis and meetingy-axis at D,(0, 3000). So region containing (0,0) represents the soln set of y \u2264 3000.
\nSince (0, 0) lies on y \u2264 3000.
\nFor region x + y \u2265 3500 : The line x + y = 3500 meets coordinate axes at E1<\/sub> (3500, 0) and F1<\/sub> (0, 3500) since (0,0) does not satisfies and hence region not containing (0,0) represents the solution set of inequation, x, y > 0 represents the first quadrant.
\nThe line x = 4500 meets line x + y = 7000 at I1<\/sub> (4500, 2500) and the line y = 3000 meets the line x + y = 7000 at H,(4000, 3000)
\nThe line y = 3000 meet x + y = 3500 intersects at G, (500, 3000)
\nThe shaded area E1<\/sub>C1<\/sub>I1<\/sub>H1<\/sub>G1<\/sub> gives the feasible region.
\nThe corner points of the feasible region are E1<\/sub>(3500,0), C1<\/sub>(4500,0), I1<\/sub>(4500, 2500), H1<\/sub>,(4000, 3000) and G1<\/sub>(500, 3000).
\n\"ML
\nThe value of Z at these corner points are as follows :<\/p>\n\n\n\n\n\n\n\n\n
Corner point<\/td>\nZ = 0.3x + 0.1y + 3950<\/td>\n<\/tr>\n
E1<\/sub>(3500,0)<\/td>\n5000<\/td>\n<\/tr>\n
C1<\/sub>(4500,0)<\/td>\n5300<\/td>\n<\/tr>\n
I1<\/sub> (4500, 2500)<\/td>\n5550<\/td>\n<\/tr>\n
H1<\/sub> (4000, 3000)<\/td>\n5450<\/td>\n<\/tr>\n
G1<\/sub> (500,3000)<\/td>\n4400 \u2192 Minimum<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The minimum value of Z is 4400 at (500, 3000).
\nThus, the oil supplied from depot A is 500 L, 3000 L, and 3500 L and from depot B is (4500 – 500) = 4000 L, (3000 – 3000) = 0 L, and (3500 – 3500) L = 0 L to petrol pumps, D, E and F respectively. The minimum transportation cost is \u20b9 4400.<\/p>\n

\"ML<\/p>\n

Question 29(Old).
\nDavid wants to invest atmost \u20b9 12000 in Bonds A and B. According to the rule, he has to invest atleast \u20b9 2000 in Bond A and atleast \u20b9 4000 in Bond B. If the rates of interest on Bonds A and B respectively are 8% and 10% per annum, formulate the problem as L.P.P. and solve it graphically for maximum interest. Also determine the maximum interest received in a year.
\nAnswer:
\nLet David invest \u20b9 x in bond A and \u20b9 y in bond B.
\nIt is given that, rates of interest on bonds A and B respectively are 8% and 10% per annum.
\nSo Total interest gained by david = \\(\\frac{8 x}{100}+\\frac{10 y}{100}\\)
\nLet Z be the total interest gained by David
\nThen Z = \\(\\frac{8 x}{100}+\\frac{10 y}{100}\\) and we want to maximise Z.
\nSince David wants to invest atmost \u20b9 12000 in bonds A and B. Then investment constraint is given by x + y \u2264 12000
\nSince, David has to invest atleast \u20b9 2000 in bond A and atleast ? 4000 in bond B. x \u2265 2000 and y \u2265 4000
\nThus the mathematical modelling of given LPP is as under :
\nMax Z = \\(\\frac{8 x}{100}+\\frac{10 y}{100}\\)
\nSubject to constraints ; x + y \u2265 12000 ;
\nx \u2265 2000 ; y \u2265 4000
\nx, y \u2265 0 [since investment can\u2019t be negative]
\nTo solve LPP, we convert inequations into equations :
\nx + y = 12000 ; x = 2000 ;y = 4000
\nFor region x + y \u2264 12000 ; The line x + y = 12000 meets coordinate axis at A (12000, 0) and B(0, 12000). Since (0, 0) lies on it. So the region containing (0, 0) gives the soln. set given inequation.
\nFor region x \u2265 2000 ; The line x = 2000 meets x-axis at C (2000, 0) and (0, 0) does not lies on x \u2265 2000 and region not containing (0, 0) represents the soln. set of x > 2000.
\nFor region y \u2265 4000 ; The line y = 4000 to x-axis meets y-axis at D (0, 4000) and region not containing (0, 0) gives the soln. set of given inequation since (0, 0) does not lies on y \u2265 4000.
\nx \u2265 0, y \u2265 0 represents the first quadrant of XOY plane.
\nThe lines x = 2000 and x + y = 12000 meets at P (2000, 10000)
\nThe lines y = 4000 and x + y = 12000 at Q (3000, 4000)
\nboth lines x = 2000 and y = 4000 intersects at R (2000, 4000)
\nWe evaluate Z at these corner points.
\n\"ML<\/p>\n\n\n\n\n\n\n
Corner point<\/td>\nZ = \\( \\frac{8 x}{100}+\\frac{10 x}{100} \\)<\/td>\n<\/tr>\n
P(2000, 1000)<\/td>\n\\( \\frac{8}{100} \\) x 2000 + \\( \\frac{10}{100} \\) x 1000 = 1160 (Max)<\/td>\n<\/tr>\n
Q(8000, 400)<\/td>\n\\( \\frac{8}{100} \\) x 800 + \\( \\frac{10}{100} \\) x 400 = 1040<\/td>\n<\/tr>\n
R(2000, 4000)<\/td>\n\\( \\frac{8}{100} \\) x 200 + \\( \\frac{10}{100} \\) x 4000 = 560<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u2234 Zmax<\/sub> = 1160 at x = 2000 ; y = 10,000
\nHence, Required investment in bond A = ? 2000 Required investment in bond B = \u20b9 10000 and Max. Interest = \u20b9 1160<\/p>\n

Question 30(Old).
\nA farmer mixes two brands P and Q of cattle feed. Brand P costing \u20b9 250 per bag, contains 3 units of nutritional element A, 2-5 units of element B and 2 units of element C. Brand Q costing \u20b9 200 per bag contains 1.5 units of nutritional element A, 11.25 units of element B and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 4.5 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost. (NCERT)
\nAnswer:
\nLet x be the no. of bags of brand P andy be the no. of bags of brand Q of cattle feed are required to minimize the cost.
\nSince costing of brand P and brand Q per bag be \u20b9 250 and \u20b9 200 respectively.
\nSo cost of x units of bags of brand P andy meets of bags of bag Q be 250x and 200y.
\nTotal cost of cattle feed be 250x + 200y
\nLet Z be the total cost of cattle feed then Z = 250x + 200y.
\nSince brand P and Q contains 3 units and 1.5 units of nutritional element A and minimum requirement of nutrient A be 18 units. Then nutrient A constraint is given by 3x + 1.5y > 18 \u21d2 2x + y \u2265 12
\nSimilarly brand P and Q contains 25 units and 11.25 units of element B and minimum requirement 4.5 units of element B be
\nSecond Constraints is given by 2.5x + 11.25y
\nSimilarly brand P and Q contains 2 unit and 3 units of element C and minimum requirement of element C be 24 units. Thus 3rd constraint is given by 2x + 3y> 24 Hence the mathematical modelling of given LPP is as under :
\nMax Z = 250x + 200y Subject to constraints ;
\n2 x + y \u2265 12 1 Ox + 45y \u2265 18
\n2x + 3y \u2265 24 and x, y \u2265 0 [since bags of cattle feed can\u2019t be negative]<\/p>\n

To solve LPP, we convert inequations into equations :
\n2x + y = 12 ;
\n10x + 45y = 18 ;
\n2x + 3y = 24 ;
\nx = y = 0
\nFor region 2x + y \u2265 12 ;
\nThe line 2x+ y= 12 meets coordinate axis at A (6, 0) and B (0, 12)
\nSince (0, 0) does not lies on given region.
\nHence region not containing (0, 0) gives the soln. set of given inequation.
\nFor region 10x + 45y \u2265 18 ;
\nThe line 10x + 45y = 18 meets coordinate axis at C (\\(\\frac{9}{5}\\), 0) and D(0, \\(\\frac{18}{45}\\)). So region not containing (0, 0) gives the soln. set of given ineqn. Since (0, 0) does not lies on it.
\nFor region 2x + 3y \u2265 24 ; The line 2x + 3y = 24 meet coordinate axis at E (12, 0) and F (0, 8). Thus region not containing (0,0) gives the soln. set of inequation. Since (0, 0) does not lies on 2x + 3y \u2265 24. x \u2265 0, y \u2265 0 represent the first quadrant of XOY plane.<\/p>\n

The lines 2x + y = 12 and 10x + 45y = 18 intersects at P\\(\\left(\\frac{261}{40},-\\frac{21}{20}\\right)\\)
\nThe lines 10x + 45y = 18 and 2x + 3y = 24 intersects at Q\\(\\left(\\frac{171}{10},-\\frac{34}{10}\\right)\\)
\nThe lines 2x + y = 12 and 2x + 3y = 24 intersects at R (3, 6).
\nThe shaded region ERB be the feasible region and is unbounded.
\nWe evaluate Z at these points E (12, 0), R (3, 5) and B (0, 12).
\n\"ML<\/p>\n\n\n\n\n\n\n
Corner point<\/td>\nZ = 250x + 200y<\/td>\n<\/tr>\n
E(12, 0)<\/td>\n250 \u00d7 12 = 3000<\/td>\n<\/tr>\n
R(3, 6)<\/td>\n250 \u00d7 3 + 200 \u00d7 6 = 1950 (Min)<\/td>\n<\/tr>\n
B(0, 12)<\/td>\n250 \u00d7 0 + 200 \u00d7 12 = 2400<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Now smallest value Z = 1950. Now we check whether this value of Z is minimum or not. For this we draw the line 250x + 200y = 1950 and check whether the open half plane 250x + 200y < 1950 have common points in feasible region or not. Here open half plane have no common points in feasible region. So smallest value is the minimum value.
\n\u2234 Zmin<\/sub> = 1950 at x = 3 and y = 6
\nrequired no. of bags of brand P = 3
\nand required no. of bags of brand Q = 6
\nand minimise cost = \u20b9 1950<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing ML Aggarwal Class 12 Solutions Chapter 3 Linear Programming Ex 3.2 is the ultimate need for students who intend to score good marks in examinations. ML Aggarwal Class 12 Maths Solutions Section C Chapter 3 Linear Programming Ex 3.2 Solve the following (1 to 12) linear programming problems graphically : Question 1. Minimize Z …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6768],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/162088"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=162088"}],"version-history":[{"count":12,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/162088\/revisions"}],"predecessor-version":[{"id":162252,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/162088\/revisions\/162252"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=162088"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=162088"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=162088"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}