ML Aggarwal Class 12 Maths Solutions Section C Chapter 3 Linear Programming Ex 3.2

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 = – 3x + 4y subject to the constraints x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.
Answer:
Draw the straight lines x + 2y = 8
and 3x + 2y = 12
x, y ≥ 0
eqn. (1) meet x-axis and y-axis at A (8, 0) and B (0, 4).
eqn. (2) meet x-axis and y-axis at C (4, 0) and D (0, 6).
eqn. (1) and eqn. (2) intersect at E (2, 3).

The shaded position OCEB is the feasible region and is bounded.
We use corner point method to find the min. value of Z, where Z = – 3x + 4y

Corner point	Value of Z
O (0, 0)	0
C (4, 0)	-3 × 4+ 4 × 0 = -12 (Min)
E (2, 3)	-3 × 2 + 4 × 3 = 6
B (0, 4)	-3 × 0 + 4 × 4 = 16

∴ Z_mim = – 12 at corner point C (4, 0).

Question 1(Old).
Maximize Z = 3x + 4y, subject to the constraints x + y ≤ 4, x ≥ 0, y ≥ 0.
Answer:
Given Max Z = 3x + 4y
Subject to constraints ; x + y ≤ 4
x, y ≥ 0
Converting the inequations into eqn’s : x + y = 4

For region x + y < 4 ; The line x + y = 4
meet 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.
The 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).

The value of objective function at these corner points is given below :

Corner point	Value of objective function Z = 3x + 4y
O (0, 0)	0
A (4,0)	12
B (0, 4)	16 (Maximum)

∴ Z_max = 16 at x = 0 and y = 4

Question 2.
Minimize Z = 3x + 5y subject to x + 3y ≥ 3, x + 3y > 2, x ≥ 0, y ≥ 0.
Answer:
Draw the straight lines x + y = 2 ………….(1)
and x + 3y = 3 …………..(2)
x, y ≥ 0
eqn. (1) meet x-axis and .y-axis at A (2, 0) and B (0, 2).
and eqn. (2) meet x-axis and^-axis at C (3, 0) and D (0, 1).
eqn. (1) and (2) intersects at E
The shaded portion represents the feasible region determined by given constraints. Clearly the region CEB is feasible and unbounded.
We calculate Z = 3x + 5y at corner points B (0, 2) ; C (3, 0) ; E (3/2, 1/2) :

Corner point	Value of Z
B(0, 2)	0 + 10 = 10
C(3, 0)	9 + 0 = 9
E(\(\frac{3}{2}\), \(\frac{1}{2}\))	\( \frac{9}{2}+\frac{5}{2} \) = 7

From 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.
For this, we draw 3x + 5y < 7. Since the open half plane has no common points with the feasible region.
So the smallest value be the minimum value.
∴ Z = 3x + 5y has no max. value and Z_min = 7 at x = \(\frac{3}{2}\) and y = \(\frac{1}{2}\).

Question 3.
Maximize Z = 4x + y, subject to the constraints x + y ≤ 50, 3x + y ≤ 90, x ≥ 0, y ≥ 0
Answer:
Draw the straight lines x+y = 50 …………..(1)
and 3x + y = 90 …(2)
x, y ≥ 0
eqn. (1) meets coordinates axis at A(50, 0) and B(0, 50)
eqn. (2) meet x-axis and y-axis at C(30, 0) and D (0, 90).
eqns. (1) and (2) intersects at E (20, 30).
Now x = 0 is y-axis and y = 0 is x-axis.
The shaded portion represents the feasible region determined by given constraints.
Clearly the region OCEB is feasible and bdd.
We find max. value of Z using corner point method
where Z = 60x + 15

The coordinates of O, C, E and B are (0, 0), (30, 0), (20, 30) and (0, 50) respectively.

Corner point	Value of Z
O(0, 0)	60 × 0 + 15 × 0 = 0
C(30, 0)	60 × 30 + 15 × 0 = 1800 (Max)
E(20, 30)	60 × 20 + 15 × 30 = 1650
B(0, 50)	60 × 0 + 15 × 50 = 750

∴ Z_max = 1800 at point C(30, 0)

Question 3(Old).
Maximize Z = 3x + 2y subject to the constraints x + 2y ≤ 10, 3x + y ≤ 15, x ≥ 0, y ≥ 0
Answer:
Draw the given st. lines
x + 2y = 10
and 3 x + y = 15
x, y ≥ 0
eqn. (1) meet x axis and y-axis at A (10, 0) and B (0, 5)
and eqn. (2) meet x-axis andy-axis at C (5, 0) (ii) D (0, 15)
Both eqn. (1) and (2) intersects at E (4, 3).
Also x = 0 is y-axis and y = 0 is x-axis.
Thus the shaded portion represents the feasible region determined by given constraints.
Clearly OCEB is feasible and bounded.
so we calculate maf value of Z using corner point method, where Z = 3x + 2y
The coordinates of O, C, E and B are (0, 0), (5, 0), (4, 3) and (0, 5) respectively.

Corner point	Value of Z
O (0, 0)	0
C (5, 0)	3 × 5 + 1 × 0 = 15
E (4, 3)	3 × 4 + 3 × 2 = 18
B (0, 5)	3 × 0 + 2 × 5 = 10

Max. Z = 18 at point E (4, 3).

Question 4.
Minimize Z = 200x + 500p, subject to the constraints x + 2y ≥ 10, 3x + 4y ≤ 24, x ≥ 0, y ≥ 0.
Answer:
Draw the straight lines x + 2y = 10 …(1) ‘
and 3x + 4y = 24 …(2) and x, y ≥ 0
eqn. (1) meets coordinate axes at A(10,0) and B(0, 5).
eqn. (2) meets coordinate axes at C(8, 0) and D (0, 6).
eqn. (1) and (2) intersects at E (4, 3).

The shaded position BED is the feasible region and is bounded.
To determine min. value of Z, we use corner point method.
where Z = 200x + 500y

Corner point	Z
B (0, 5)	200 x 0 + 500 x 5 = 2500
E (4, 3)	200 x 4 + 500 x 3 = 2300 (Min)
D (0, 6)	200 x 0 + 500 x 6 = 3000

∴ Min. Z = 2300 at E (4, 3).

Question 5.
Minimize Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, y ≥ 0. Show that the minimum value of Z occurs at more than two points.
Answer:
Converting the given inequations into equations :
2x + y = 3
x + 2y = 6
x = y = 0
The 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.

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.
Region x > 0, y > 0 ; It represents the first quadrant of XOY plane.
Also both lines 2x + y – 3 and x + 2y = 6 intersects at P (0, 3)
So the shaded area CP gives the feasible region and it is unbounded with corner points C (6,0) and P (0, 3).

Corner point	Z = x + 2y
C (6, 0)	6
P (0, 3)	6

Z 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.

Question 6.
Minimize and maximize Z = 3x + 9y subject to the constraints x + y ≥ 10, x + 3y ≤ 60, x ≤ y, x ≥ 0, y ≥ 0. (NCERT)
Answer:
Draw the straight lines x + 3y = 60 …(1)
x + y = 10 …(2)
and x = y ..(3)
eqn. (1) meet x-axis and y-axis at A (60, 0) and B (0, 20).
eqn. (2) meet x-axis and y-axis at C(10, 0) and D (0, 10)
eqn. (3) passes through (0, 0).
eqn. (3) meets eqn. (1) at M (15, 15) and meets eqn. (2) at N(5, 5).
The shaded portion represents the feasible region determined by given constraints.
Also eqn. (1) and (2) intersects at P (- 15, 25).
Clearly the feasible region DNMB is bounded.
We use corner point method to find the max. and min value of Z where Z = 3x + 9y

Corner point	Value of Z
D(0, 10)	3 × 0 + 9 × 10 = 90
N (5, 5)	3 × 5 + 9 × 5 = 60 (Min)
M (15, 15)	3 × 15 + 9 × 15 = 180
B (0, 20)	3 × 0 + 9 × 20 = 180 (Max)

∴ Min Z = 60 at point (5, 5) and Max. Z = 180 at the points (15, 15) and (0, 20).
i.e. Maximum value of Z occurs at all points on the line joining M and B.

Question 7.
Minimize Z = 3x + 2y subject to the constraints x + y ≥ 8, 3x + 5y ≤ 15, x ≥ 0, y ≥ 0. (NCERT)
Answer:
Converting the given inequations into equations :
-x + y = 8 ;
3x + 5y= 15 ;
x = 0 = y
For region x + y ≥ 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 ≥ 8.
Thus, region not containing (0, 0) gives the soln. set of given inequation.
For Region 3x + 5y ≤ 15 ;
The line 3x + 5y = 15 meets coordinate axis at C (5, 0) and D (0, 3).
Since (0, 0) lies on 3x + 5y ≤ 15.
Thus the region containing (0, 0) gives the solution set of given inequation.
For region x, y ≥ 0 ; It represents the first quadrant of xoy plane.
Also both lines intersects at P\(\left(\frac{25}{2},-\frac{9}{2}\right)\)
Clearly their is no common feasible region.

Question 8.
Minimize and mximize Z = x + 2y subject to the constraints
x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200, x ≥ 0, y ≥ 0. (NCERT)
Answer:
Draw the st. lines x + 2y= 100 …(1)
2x – y = 0 …(2)
and 2x + y = 200 ……….(3)
x, y ≥ 0
eqn. (1) intersects x-axis and 7-axis at A (100, 0) and B (0, 50)
(2) line meets x-axis and 7-axis at C (100, 0) and D (0, 200)
eqn. (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).
The shaded region BMND is feasible and bounded we determine the max or min. value of Z = x + 2y using corner point method.
Z = x + 2y

Corner point	Z = x + 2y
B(0, 50)	0 + 2 × 50 = 100
M(20, 40)	20 + 2 × 40 = 100
N(50, 100)	50 + 2 × 100 = 250
D(0, 200)	0 + 2 × 200 = 400 (Max)

∴ Max. Z = 400 at the point (0, 200).
and Min. Z = 100 at the points (0, 50) and (20, 40) i.e. line segment joining B and M.

Question 9.
Maximize Z = x + y subject to the constraints x – y ≤ -1, -x + y ≤ 0, x ≥ 0, y ≥ 0. (NCERT)
Answer:
We draw the straight lines x – y = – 1 …(1)
and -x + y = 0 ……….(2)
x, y ≥ 0
line (1) meet x-axis andy-axis at A (- 1, 0) and B (0, 1)
line (2) meet x-axis and y-axis at C (0, 0)
(1) and (2) intersects at no point.
Clearly there is no point which satisfies all constraints simultaneously.
Thus there is no feasible region and hence there is no feasible solution.

Hence maximum value of Z does not exists.

Question 10.
A 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 ₹ 17.50 per package on nuts and ₹ 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.
Answer:
The given information can be summarised in tabular form is given as under :

Time required to produce products
Machines	Nuts	Bolts	Max. machine hours available
A	1	3	12
B	3	1	12
Profit (in ₹)	₹ 17.50	₹ 7

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.
Thus, x + 3y ≤ 12
Similarly for machine, the constraint be given as; 3x +y ≤ 12
Now it is given that, profit of ₹ 17.50 per package on nuts and ₹ 7 per package of bolts.
Thus total profit on x-package of nuts and y-package of bolt be 17.50x + 7y.
Let Z be the total profit. Then Z = 17.50x + 7y.
Clearly x ≥ 0 and y ≥ 0
Thus the given LPP can be stated mathematically as;
x + 3y ≤ 12
3 x + y ≤ 12
x ≥ 0, y ≥ 0
and Max Z = 17.50x + 7y
For 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 ≤ 12 [∵ 0 ≤ 12] Thus, region containing (0, 0) gives the soln. set of x + 3y ≤ 12 For region 3x + y ≤ 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.
The lines x + 3y = 12 and 3x + y = 12 intersect at P (3, 3).
Now 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).

The values of objective function at these comer points is given as under :

Corner points	Z = 17.50x + 7y
O(0, 0)	Z = 0
C (4, 0)	Z = 17.50 × 4 + 7 × 0 = 70
P (3, 3)	Z = 17.50 × 3 + 7 × 3 = 73.50
B (0, 4)	Z = 0 + 7 × 4 = 28

Clearly Z is maximum at x = y = 3 and Z_max = ₹ 73.50
Here the manufacturer produces 3 packages of each nuts and bolts per day to realise maximum profit of ₹ 73.50.

Question 10.
Maximize Z = -x + 2y subject to the constraints x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0. (NCERT)
Answer:
Converting the given inequations into equations. x = 3 ;x + y = 5 ; x + 2y = 6 ; y = 0
For region x ≥ 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 ≥ 3. So region not containing (0, 0) gives the solution set of x > 3.
For 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 ≥ 5. Thus region not containing (0, 0) gives the solution set of given inequation.
For region x + 2y ≥ 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 ≥ 6. Thus, the region not containing origin gives the soln. set of given inequation.
Region x, y ≥ 0 represents the first quadrant of XOY plane.
The lines x = 3 and x + y = 5 intersects at P (3, 2).
The lines x = 3 and x + 2y = 6 intersects at Q(3, \(\frac{3}{2}\)).
The lines x + y = 5 and x + 2y = 6 intersects at R (4, 1).
The shaded position PRD gives the feasible region.
So min value of Z exists but maximum value of Z does not exists since feasible region is unbounded.

Question 11.
A 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 ?
Answer:
Let the required number of gold rings and chains manufactured by a firm are x and y respectively to get the maximum profit.
Since, profit on each ring and chains are ₹ 300 and ₹ 190 respectively, so, profit onx units of rings and y units of chains are ₹ 300x and ₹ 190y respectively
Let Z be total profit. Then Z = 300x + 190y and we have to maximize.
Since 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 × 60 = 960, so (first constraint) is given by .
60x + 30y ≤ 960 Thus, 2x + y ≤ 32
Given, total number or rings and chains manufactured is at most 24, so, second constraint is given by x + y ≤ 24
Hence, the mathematical formulation of given LPP is given below:
Maximum Z = 300x + 160y
Subject to constraints,
2x + y ≤ 32 x + y ≤ 24
x, y ≥ 0 [Since production can not be less than zero]
Region 2x +y ≤ 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 ≤ 32 as (0,0) satisfies 2 x + y ≤ 32. Region x + y ≤ 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 ≤ 24 as (0,0) satisfies x + y ≤ 24. Region x, y ≥ 0 : it represents the first quadrant of xy-plane.
Thus Shaded region OCEB represents feasible region.

The point E(8, 16) is obtained by solving 2 x+y = 32 and x +y = 24 simultaneously.

Corner Point	Z = 300x + 160y
0(0,0)	300(0) + 160(0) = 0
C(16,0)	300(16) + 160(0) = 4800
E(8, 16)	300(8) + 160(16) = 4960
B(0,24)	300(0) + 160(24) = 3840

Z_max or maximum Z = 4960 at x = 8, y = 16
reqd. number of rings = 8 & reqd. no. of chains = 16 & maximum profit = ₹ 4960

Question 12.
A 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.
Answer:
Let 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.
Let Z be the total profit of company.
Then Z = 40x + 50y
We make the given data in tabulated form as under :

goods	silver (in gm)	gold (in gm)	Requirement
A	3	1
B	1	2
Requirement	9	8

since it is given that, the company can use atmost 9 gm of silver ∴ 3x + y < 9
Also, it is given that, the company can use atmost 8 gm of gold ∴ x + 2y < 8
Thus the mathematical modeling of given LPP is as under :
Max Z = 40x + 50y
Subject to constraints : 3x + y ≤ 9; x + 2y ≤ 8 ; x ≥ 0, y ≥ 0 [since no. of units of good A and B can’t be negative]
For region 3x + y ≤ 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.
For region x + 2y ≤ 8; The line x + 2y = 8 meets coordinate axes at C(8, 0) and D(0, 4). Clearly (0, 0) satisfies the given inequality. .
Thus, region containing (0, 0) gives the solution set of given inequality.
Region x ≥ 0, y ≥ 0 represents the first quadrant of XOY plane. Both lines 3x + y = 9 and x + 2y = 8 intersects at P(2, 3).
Thus 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).
We evaluate the objective function Z at these comer points.

Corner points	Z = 40x + 50y
O(0, 0)	0
A(3, 0)	40 × 3 + 50 × 0 = 120
P(2, 3)	40 × 2 + 50 × 3 = 230
D(0, 4)	40 × 0 + 50 × 4 = 200

Clearly Z_max = 230 at P(2, 3) i.e. x = 2 and y = 3
Thus, the company should produce 2 units of good A and 3 units of good B to get maximum profit Rs. 230.

Question 13.
A 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 ‘A’, while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a package of screws ‘B’ Each machine is available for atmost 4 hours on any day. The manufacturer can sell a package of screws ‘A’ at a profit of ₹ 7 and of screws ‘B’ at a profit of ₹ 10. Assuming that he can sell all the screws he can manufacture, how many packages of each type should the factory owner produce in a day in order to maximize his profit ? Determine the maximum profit. (NCERT)
Answer:
Let the factory manufactures x screws of type A and y screws of type B on each day to maxmise the profit.
Therefore, x ≥ 0 and y ≥ 0

The given information can be tabulated as given below

	Screw A	Screw B	Availability
Automatic Machine (min)	4	5	4 × 60 = 240
Hand Operated Machine (min)	6	3	4 × 60 = 240

The profit on a package of screws A is ₹ 7 and on the pakage of screws B is ₹ 10. Therefore, the constraints are;
4x + 6y ≤ 240
6x + 3y ≤ 240
Total profit, Z = 7x + 10y
The required mathematical formulation of the given problem is as follows :
Maximum Z = 7x + 10y Subject to the constraints,
4x + 6y ≤ 240
6x + 3y ≤ 240
x, y ≥ 0
Converting given inequations into equations;
4x + 6y = 240 ⇒ 2x + 3y = 120 …(1)
& 6x + 3y = 240 ⇒ 2x +y = 80 ……..(2)
and x = y = 0
For Region 2x + 3y ≤ 120 : The line 2x + 3y = 120 meets coordinate axes at A₁ (60, 0) and B₁ (0, 40). Also (0,0) satisfies 2x + 3y ≤ 120. Thus the region containing (0, 0) gives the solution set of 2x + 3y ≤ 120.
For Region 2x + y ≤ 80 : The line 2x + y = 80 meets coordinate axes at A₂ (40, 0) and B₂ (0, 80) Clearly (0,0) satisfies it. These the region containing (0, 0) gives the solution set of inequation 2x + y ≤ 80.
Region x ≥ 0; y ≥ 0 represents the first quadrant, both lines (1) & (2) intersects at P (30, 20).
The 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).
The values of Z at these corner points are as follows :

Corner point	Z = 7x + 10y
O (0,0)	0
A2(40,0)	280
P (30,20)	410 (Maximum)
B1 (0, 40)	400

The maximum value of Z is 410 i.e. Z_max = 410 at (30, 20).
Hence, the factory should produce 30 package of screws A and 20 packages of screws B to get the maximum profit of ₹ 410.

Question 13(Old).
A 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.
Do 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 ?
Answer:
Let 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 :
Maximize Z = 100x + 120y Subject to constriants ;
2x + 3y ≤ 30
3x + y ≤ 17
and x ≥ 0, y ≥ 0
To solve the LPP we convert the inequations into equations :
2x + 3 y = 30
x + y = 17
x = 0 = y
For region 2x + 3y ≤ 30 ; The line 2x + 3y = 30 meets the coordinate axes at A₁ (15, 0) and A (0, 10). Clearly (0, 0) satisfies the inequation 2x + 3y ≤ 30. Hence the region containing (0, 0) gives the solution set of given inequation.
For region 3x + y ≤ 17 ; The line 3x + y = 17 meets coordinate axes at A₂ (17/3, 0) and B₂ (0, 17). Clearly (0, 0) satisfies 3x + y ≤ 17. Hence the region containing (0, 0) gives the solution set of given inequation.
Further, x ≥ 0, y ≥ 0 represents the first quadrant of xoy plane.
Both lines 2x + 3y = 30 and 3x + y = 17 intersects at C(3, 8).
The shaded area gives the feasible region OA₂PB₁

The coordinates of the vertices (Corner – points) of shaded feasible region OA₂CA are O (0, 0), A₂(\(\frac{17}{3}\), 0), C(3, 8) and A(0, 10)
The values of the objective of function at these comer points are given in the following table :

Corner Point	Value of objective function Z = 100.x + 120y
A2(\(\frac{17}{3}\), 0)	Z = 100 × y+ 120 × 0 = 566. 67
C(3, 8)	Z = 100 × 3 + 120 × 10 = 1260
A(0, 10)	Z = 100 × 0 + 120 × 10 = 1200
O(0, 0)	Z = 0

Clearly Z_max = 1260 at x = 3 and y = 8
Hence, 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 ₹ 1260.
Yes, because efficiency of a person does not depend on sex (male or female).

Question 14.
A dealer in rural area wishes to purchase a number of sewing machines. He has only ₹ 57600 to invest and has space for atmost 20 items. An electronic sewing machine costs him ₹ 3600 and a manually operated sewing machine ₹ 2400. He can sell an electronic sewing machine at a profit of ₹ 220 and a manually operated sewing machine at a profit of ₹ 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.
Answer:
Let x and y are the number of electronic and manually operated sewing machine that a dealer buys and sells.
The dealer gets a profit of ₹ 220 and ₹ 180 on selling on electronic and manually operated sewing machine. .
So Total profit of dealer = 220x + 180y
Let Z be the total profit of dealer.
Since cost of 1 electronic and 1 manually operate sewing machine are ₹ 3600 and ₹ 2400 respectively.
Thus cost of x electronic and y manually operated sewing machines are 3600x and 2400y. also the dealer can investment atmost ₹ 57600.
Thus purchased constraint is given by 3600x + 2400y ≤ 57600 ⇒ 3x + 2y ≤ 48
Also the dealer has a space to store atmost 20items.
Thus storage constraint is given by x + y ≤ 20
Hence, the mathematical formulation of given LPP is given as under:
Max Z = 220x + 180
Subject to Constraints ; 3x + 2y ≤ 48
x + y ≤ 20
x ≥ 0 ; y ≥ 0
[Since electronic and manually operated sewing machines can’t be negative]
To solve LPP, we convert inequations into eqn’s 3x + 2y = 48; x + y = 20; x = 0 = y
For region 3x + 2y ≤ 48 ; The line 3x + 2y = 48 meets coordinate axis at A (16, 0) and B (0, 24).
Since (0, 0) lies on given inequation. Thus, region containing (0, 0) gives soln. set of 3x + 3y ≤ 48. For region x + y ≤ 20 ;
The line x + y = 20 meets coordinate axis at C (20, 0) and D (0, 20).
So region containing (0, 0) gives the soln. set of given inequation since (0, 0) lies on x + y ≤ 20.
The region x ≥ 0, y ≥ 0 represents the first quadrant of XOY plane.
Both lines intersects at P (8, 12).
Thus 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).
We evaluate Z at these corner points

Corner point	Z = 220x + 180y
O (0, 0)	0
A (16, 0)	220 × 16 = 3520
P(8, 12)	220 × 8 + 180 × 12 = 3920 (Max)
D(0, 20)	220 × 0 + 180 × 20 = 3600

∴ Z_max = 3920 at P (8, 12) i.e. at x = 8 ; y = 20
required no. of electronic sewing machines = 8
required no. of normally operated machines = 12
and Max. profit = ₹ 3920
Keeping in view of saving environment and conservation of exhausitible resources, the manually operated machines should be promoted.

Question 14 (Old).
A 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 ₹ 50 and each unit of item Q makes a profit of ₹ 60, determine the number of units of each item that the company should produce to maximize profit. What is the maximum profit? (ISC 2010)
Answer:
Let 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 ₹ 50 and ₹ 60 respectively. So, profits on x goods of type P and y goods of type Q are 50x and 60y respectively,
Let Z be total profit on A and B.
Then Z = 50x + 60y
Since, 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
3x + y ≤ 9
Since, 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
x + 2y ≤ 8
Hence mathematical formulation of LPP is, given below:
Max Z = 50x + 60y
Subject to constraints,
3 x + y ≤ 9
x + 2y ≤ 8
x, y ≥ 0 [Since products of A and B can not be less than zero]
Region 3x + y ≤ 9 : Then line 3x + y = 9 meets coordinate axes at A₁(3, 0) & B₁(0,9) respectively.
Region containing origin respresents the solution set of 3x + y ≤ 9 as (0,0) satisfies 3x + y ≥ 9.
Region x + 2y ≤ 8 : line x + 2y = 8 meets coordinate axes at C₁(8,0) & D,(0,4) respectively.
Region containing origin represents the solution set of x + 2y ≤ 8 as (0,0) satisfies x + 2y ≤ 8.
Region x, y ≥ 0 : it represents first quardant of xy region.
Shaded region OA₁E₁ is the feasible region.
The points E₁(2,3) is obtained by solving 3x + y = 9 and x + 2y = 8 simultaneously.
The corner points of feasible region are O(0, 0); A₁(3, 0); E₁(2, 3) and D₁(0, 40).

Corner point	Z = 50x + 60y
O(0,0)	50(0) + 60(0) = 0
A(3,0)	50(3) + 60(0)= 150
E1(2,3)	50(2) + 60(3) = 280
D1(0, 40)	50(0) + 60(4) = 240

Therefore maximum Z = 280 at x = 2, y = 3

Hence, reqd. Maximum profit = ₹ 280 and required number of goods of type A = 2, and reqd no. of goods of type B = 3.

 

Question 15.
A 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 ₹ 5 each for type A and ₹ 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize the profit ? (NCERT)
Answer:
Let the company manufacture x souvenirs of type A and y souvenirs of type B.
Therefore, x ≥ 0 and y ≥ 0
The given information can be put in tabular form is follows as :

	Type A	Type B	Availability
Cutting (min)	5	8	3 x 60 + 20 = 200
Assembing (min)	10	8	4 x 60 = 240

The profit on type A souvenirs is ₹ 5 and on type B souvanirs is ₹ 6.
Total profit, Z = 5x + 6y
The required mathematical formulation of the given problem is Maximise Z = 5x + 6y
subject to the constraints,
5x + 8y ≤ 200
5x + 4y ≤ 120
x, y ≥ 0
For solution of LPP, We convert inequations into equations
5x + 8y = 200
& 5x + 4y = 120
& x = 0 = y

For region 5x + 8y ≤ 200 : The line 5x + 8y = 200 meet coordinate axes at A₁(40,0) and B₁(0, 25). so region containing (0,0) be the solution set of inequation 5x + 8y < 200 as (0,0) lies on 5x + 8y ≤ 200. For region 5x + 4y ≤ 120 : The line 5x + 4y = 120 meet coordinate axes at A₂(24, 0) and B₂(0, 30) and region contaning (0,0) represents the solution set of inequation 5x + 4y ≤ 120 as (0, 0) satisfies the inequation.Now region x ≥ 0; y ≥ 0 gives the first quadrant of xy plane.
Both lines intersects at P (8, 20).
The feasible region OA₂PB₁ determined by the system of constraints is as follows.
The corner points are O(0,0), A₂ (24,0) P (8, 20) and B₁ (0, 25).
The values of Z at these corner points are as follows.

Corner point	Z = 5x + 6y
O(0,0)	0
A2(24, 0)	= 5 × 24 + 6 × 0=120
P(8, 20)	= 5 × 8 + 6 × 20 =160
B1(0, 25)	150 – Maximum

The maximum value of Z is 160 at (8, 20) i.e. x = 8 and y = 20
Thus, 8 souvenirs of type A and 20souvenirs of type B should be produced each day to get the maximum profit of ₹ 160.

Question 15 (Old).
A cooperative society of farmers has 50 hactare of land to grow two crops X and Y. The Pr°fit from crop X and Y per hectare are estimated as ₹ 10500 and ₹ 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.
Keeping 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 ?
Answer:
Let x hectares of land grows crop X. and y hectares of land grows crop Y.
Then the mathematical modelling of the LPP is given as follows :
Maximize Z = 10,500x + 9,000y
Subject to constraints ; x + y ≤ 50,
20x + 10y ≤ 800
and x ≥ 0, y ≥ 0
To solve the LPP we convert inequation into equations x + y = 50
20x + 10y = 800 ; x = y = 0
For region x + y ≤ 50 ; The line x + y = 50 meets coordinate axes at A₁(50, 0) and B₁ (0, 50).
Since (0, 0) satisfies given in equation so the region containing (0, 0) gives the solution set of x + y < 50.
For region 20x + 10y < 800 ;
The line 20x + 10y = 800 meets coordinate axes.at A₂(40, 0) and B₂(0, 80).
The region containing (0, 0) gives the solution set of inequation since (0, 0) satisfies the inequation.
x ≥ 0 ; y ≥ 0 represents the first quadrant, both lines intersects at P (30, 20).
Thus the shaded area OA₂PB₁ be the feasible region.
The feasible region of the LPP is shaded in graph.

The coordinates of the vertices (Corner – points) of shaded feasible region OA₂PB₁ are O (0, 0), A₂(40, 0), P (30, 20) and B₁(0, 50).
The values of the objective of function at these points are given in the following table :

Point (x2, y2)	Value of objective function Z = 10,500x + 9,000y
A2 (40, 0)	Z = 10500 × 40 = 4,20,000
P (30, 20)	Z = 10500 × 30 + 9000 × 20 = 4,95,000
B1 (0, 50)	Z = 10500 × 0 + 9000 × 50 = 4,50,000
O (0, 0)	Z = 0

Here Z_max = 4,95,000 at x = 30 ; y = 20
Thus, 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.
Thus, the required maximum profit that can be earned is ₹ 4,95,000.
Yes, I agree with the message that the protection of wildlife is utmost necessary to preserve the balance in environment.

Question 16.
A 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 ₹ 7 and B at a profit of ₹ 4, find the production level for maximum profit graphically. (Type CBSE Delhi 2016)
Answer:
Given data can be arranged in tabular form is as under :

Product Machine	I	II	Profit (in ₹)
A	3	3	7
B	2	1	4
Maximum availability capacity	12	9

Let x units of product A and y units of product B are produced by manufacturer.
Since it is given that each unit of product A is sold at profit of ₹ 5 and that of product B is sold at profit of ₹ 6. Thus the total profit earned by manufacturer is selling x units of product A and y units of product B be 5x + 6y.
Let Z be the total profit ∴ Z = 7x + 4y and our aim is to maximise Z. and corresponding constraints are : 3x + 2y ≤ 12 (Machine I constraint)
3x + y ≤ 9 (Machine II constraint)
x ≥ 0; y ≥ 0
[since no. of units of product A and B can’t be negative]
For region 3x + 2y ≤ 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.
For region 3x + y ≤ 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 ≥ 0, y ≥ 0 represents the first quadrant of XOY plane.
Both 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.

Corner points	Z = 7x + 4y
O (0, 0)	7 × 0 + 4 × 0 = 0
C (3, 0)	7 × 3 + 4 × 0 = 21
P (2, 3)	7 × 2 + 4 × 3 = 26
B (0, 6)	7 × 0 + 4 × 6 = 24

Z_max = 26 at B (0, 6) i.e. at x = 0 and y = 6
Hence maximum profit of ₹ 26 is earned when 0 units of product A and 6 units of product B are produced.

Question 16(Old).
One 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.
Answer:
Let x be the required number of one kind of cake and y be the required number of second kind of cakes that are made.

Let Z be the total no. of cakes. Then Z = x + y
Since 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 × 1000) then first constraint be, 2.00x + 100y < 5000
⇒ 2x + y < 50
Since 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 ≤ 5000, 25x + 50y ≤ 1000 and x ≥ 0, y ≥ 0 To solve the LPP we convert the inequations into equations : 2x + y = 50 ; x + 2y = 40 ; x = 0 = y For region 2x + y ≤ 50 ; The line meet coordinate axes at A(25, 0) and E(0, 50). Since (0, 0) satisfies 2x + y ≤ 50. Thus the soln. set of 2x + y ≤ 50 containing (0, 0). For region x + 2y ≤ 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).
The feasible region OABC of the LPP is shaded in graph.

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).
The value of the objective of function at these points are given in the following table:

Corner Point	Value of objective function Z = x + y
A (25, 0)	Z = 25
B(20, 10)	Z = 30
C (0, 20)	Z = 20
O (0, 0)	Z = 0

The maximum of 30 cakes can be made out of which 20 are of one kind and 10 are of IInd kind.

Question 17.
A 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 :
Answer:
Let 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 ≤ 42 => x + 2y ≤ 28 Similarly constraint associated with craftsman time be given as 3x + y ≤ 24
Thus the given LPP can be represented mathematically as under: max Z = 30.x + 20y Subject to constraints; x + 2y ≤ 28 3x + y ≤ 24; x ≥ 0, y ≥ 0
For region x + 2y ≤ 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.
For region 3x + y ≤ 24 ; The line 3x + y = 24 meets coordinate axes at C(8, 0) and D (0, 24). Clearly (0, 0) lies on 3x + y ≤ 24. Thus region containing (0,0) gives the soln. set of given inequality.

Now x ≥ 0, y ≥ 0 represents the positive quadrant of XOY plane.
Both lines x + 2y = 28 and 3x + y = 24 intersects atP (4, 12)
Clearly 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).

Corner points	Z = 30x + 20y
O (0,0)	Z = o
C (8, 0)	Z = 30 × 8 + 20 × o = 240
P (4,12)	Z = 30 × 4 + 20 × 12 = 120 + 240 = 360
B (0, 14)	Z = 30 x 0 + 20 x 14 = 280

Thus Z is maximum at x = 4 and y = 12
and Z_max = 360
Hence the manufacturer produces 4 pedestal lamps and 12 wooden shades to realise max profit of ₹ 360.

Question 17(Old).
A 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 ₹ 50 each on a toy of Type A and ₹ 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)
Answer:
Let 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 ₹ 50 and ₹ 60 respectively, so, profits on x units of toys A and y units of toy B are ₹ 50x and ₹ 60y respectively. Let Z be total profit.
Then Z = 50x + 60y
Since 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
5x + 8y ≤ 180
Since 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
10x + 8y ≤ 240
i. e. 5x + 4y ≤ 120
Hence the mathematical formulation of LPP is given below :
maximize Z = 50x + 60y
Subject to constraints,
5x + 8y ≤ 180
[Since production can not be less than zero]
Region 5x + 8y ≤ 180 : The line 5x + 8y = 180 meets coordinate axes at A₁(36,0) and B₁(0, \(\frac{45}{2}\)) respectively.
Region containing origin represents the solution set of 5x + 8y ≤ 180 as (0,0) satisfies 5x + 87 ≤ 180.

Region 5x + 4y ≤ 120 : The line 5x + 4y = 120 meets coordinate axes at C₁(24, 0) and D₁(0, 30) respectively.
Region containing origin represents the solution set of 5x + 4y ≤ 120, as (0,0) satisfies 5x + 4y ≤ 120. Region x, y ≥ 0 : it represent first quadrant of xy plane.
The shaded region OC₁E₁B₁ represents feasible region.
The point E₁( 12, 15) is the point of intersection obtained by solving 5x + 8y = 180 and 5x + 4y = 120 simultaneously.

Corner Point	Z = 50x + 60y
O(0, 0)	50(0) + 60(0) = 0
C1(24, 0)	50(24) + 60(0) = 1200
E1(12, 15)	50(12) + 60(15) = 1500
B1(0, \(\frac{45}{2}\))	50(0) + 60(\(\frac{45}{2}\)) = 1350

Clearly maximum Z = 1500 at x = 12, y = 15
Thus required Number of toys A = 12, required no. of toys B = 15 & Maximum profit = ₹ 1500

Question 18.
An 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 ₹ 500 and on each economy class ticket is ₹ 300. Determine how many tickets of each type must be sold to maximize the profit of the airline. Also find the maximum profit.
Answer:
Let A andy be the number of first class and economy class tickets. Now it is given that, profit on each first class ticket is ₹ 500 and on each economy class ticket is ₹ 300.
Thus, the profit gained by airline on selling A ticket of first class and y ticket of economy class be 500x + 300y.
Let Z be the total profit of airline.
Then Z = 500A + 300y and we want to maximize Z.
The mathematical modelling of given LPP is given as follows : max Z = 500x + 300y
Subject to constriants ; x + y ≤ 200 [carry constraints]
30x + 20y ≤ 4500 [baggage constraint]
i. e. 3x + 2y ≤ 450
Also, x ≥ 0; y ≥ 0 [Since first class and economy class tickets can’t be negative]
For region x + y ≥ 200 ; The line x + y = 200 meet coordinate axis at A (200, 0) and B(0, 200). Since (0, 0) lies on x + y ≤ 200.
∴ The region containing (0, 0) gives the soln. set of given inequation.
For region 3x + 2y ≤ 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.
Region x, y ≥ 0 represents the first quadrant of XOY plane.
Both given lines intersects at P (50, 150).
Thus 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).
We evaluate Z at these corner points

Corner point	Z = 500x + 300y
O (0, 0)	0
C (150, 0)	500 × 150 + 0 = 75000
P (50, 150)	500 × 50 + 300 × 150 = 25000 + 45000 = 70000
B (0, 200)	0 + 300 × 200 = 60, 000

Z_max = 75,000 at C (150, 0) i.e. at x = 150 and y = 0
Thus, required no. of first class ticket = 150
and required no. of economy class ticket = 0
and Maximum profit = ₹ 75,000

Question 18(Old).
A manufacturer manufactures two types of plastic bags A and B using two machines M₁ and M₂. TO manufacture a bag of type A, machine M₁ is operated for 2 minutes and machine M₂ for 1 minute ; to manufacture a bag of type B, machine M₁ is operated for 3 minutes and machine M₂ 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 ₹ 0.50 and a bag of type B is sold at a profit of ₹ 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)
Answer:
Let 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.
Since it is given that, bag A is sold at a profit of ₹ 0.50 and bag B is sold at profit of ₹ 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.
Then Z = 0.50x + 0.75y and we want to maximize Z.
To manufacture bag A and B, machine M₁, is operated for 2 min and 3 minutes and machine M₁, can be operated for atmost 8 hours per day.
∴ The corresponding constraint is given by
2x + 3y < 8 x 60 = 480
To manufacture bag A and B, machine M₂ is operated for 1 minute and 4 minute and also machine M₂ can be operated for atmost 8 hrs. per day.
So, the corresponding constraint is given by x + 4y < 8 x 60 — 480
Thus, the mathematical modelling of given LPP is as under :
Max Z = 0.50+ 0.75y Subject to Constraints ;
2x + 3y ≤ 480
x + 4y ≤ 480
x, y ≥ 0 [Since no. of bags A and bag B cannot be negative]

To solve LPP, we convert inequations into eqns :
2x + 3y = 480 ;
x + 4y = 480 ;
x = 0 = y
For region 2x + 3y ≤ 480 ;
The line 2x + 3y = 480 meets coordinate axis at A (240, 0) and B (0, 160).
Since (0, 0) lies on 2x + 3y ≤ 480. So region containing origin gives the solution set of given inequation.

For region x + 4y ≤ 480 :
The line x + 4y = 480 meets coordinate axis at C(480, 0) and D(0, 120).
Since (0, 0) lies on x + 4y ≤ 480. Thus region containing (0, 0) gives the soln. set of given inequation.
x ≥ 0, y ≥ 0 represents the first quadrant of XOY plane. Both lines intersects at P (96, 96).
Thus 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).

Corner Point	Z = 0.5x + 0.75y
O (0, 0)	0
A1(240, 0)	0.5 × 240 =120
P(96, 96)	0.5 × 96 + 0.75 × 96 =120
D (0,120)	0.75 × 120 = 90

Clearly Z_max =120 at x = 240 or y = 0 or at x = y = 96.
Required 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.

Question 19.
A 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)
Answer:
Let 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.
It is given that product A is sold for Rs. 48 per unit and product B is sold for Rs. 40 per unit.
Let Z be the maximum profit of carpenter Then Z = 48x + 40y
The given data of given LPP is tabulated as under :

The mathematical modeling of given LPP is given as under Max Z = 48x + 40y
Subject to constraints :
2x + y ≤ 90 ; x + 2y ≤ 80; x + y ≤ 50
x ≥ 0 y ≥ 0 [since no. of units can’t be negative]
For 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.
For region x + 2y ≤ 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.
For region x + y ≤ 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.
Region x ≥ 0, y ≥ 0 represents the first quadrant of XOY plane
The lines 2x + y = 90 and x + 2y = 30 intersects at P(\(\left(\frac{100}{3}, \frac{70}{3}\right)\))
The lines x + 2y = 80 and x + y = 50 intersects at Q(20, 30).
The lines 2x + y = 90 and x + y = 50 intersects at R(40, 10).
Clearly 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 :

Corner points	Z = 48x + 40y
0(0, 0)	0
A(45, 0)	48 × 45 = 2160
R(40,10)	48 × 40 + 40 × 10 = 2320 4
Q(20, 30)	8 × 20 + 40 × 30 = 2160
D(0, 40)	= 1600

Clearly Z_max = 2320 at x = 40 and y = 10.
Thus 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.

Question 20.
A diet is to contain atleast 80 units of Vitamin A and 100 units of minerals. Two foods F₁ and F₂ are available costing ₹ 5 per unit and ₹ 6 per unit respectively. One unit of food F₁ contains 4 unitsof vitamin A and 3 units of minerals whereas one unit of food F₂ 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.
Answer:
Let x units of food F₁ and y units of food F₂ respectively used to minimise the cost. Given cost of 1 unit of food F₁ and food F₂ be ₹ 5 and ₹ 6 respectively. Then cost of x units of food F₁ and y units of food F₂ be ₹ 5x and ₹ 6y respectively. Let Z be the total cost of diet.
Then Z = 5x + 6y
The given problem can be summarized in tabular form is as under :

Thus the mathematical modelling of given LPP is given as under :
Min Z = 5x + 6y
Subject to Constraints :
4x + 3y ≥ 80
3 x + 6y ≥ 100
x ≥ 0, y ≥ 0
For region 4x + 3y ≥ 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 ≥ 80.

Then Z = 50x + 70y
The given problem can be summarised in tabular form is as under :


Thus, the mathematical modelling of given LPP is given as under :
Min. Z = 50x + 70y Subject to constraints ;
2x + y ≥ 8
x + 2y ≥ 0
x ≥ 0, y ≥ 0
For region 2x + y ≥ 8;
The 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.
For region x + 2y > 10 ; The line x + 2y = 10 meets coordinate axes at C (10, 0) and D(0, 5).
Clearly (0,0) does not satisfies x + 2y > 10. Thus region not containing (0, 0) gives the soln. set of given inequality.
Further x ≥ 0, y > 0 represents the first quadrant of XOY plane.
The lines 2x + y = 8 and x + 2y = 10 intersects at P (2, 4).
The unbounded shaded region BPC represents the feasible region with comer points B (0, 8); P (2, 4) and C (10, 0).

Corner points	Z = 50x + 70y
B (0, 8)	Z = 50 × 0 + 70 × 8 = 560
P (2, 4)	Z = 50 × 2 + 70 × 4 = 380
C (10, 0)	Z = 50 × 10 + 70 × 0 = 500

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.
∴ Z_min = 380 at x = 2 andy = 4
Thus 2 kg of food I and 4 kg of food II are mixed to realize minimum cost.

Question 22.
Reshma 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)
Answer:
Let 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.
Then the required mathematical model of the LPP is given as follows :
Minimize Z = 60x + 80y Subject to contraints,
3x + 4y ≥ 8,
5x + 2y ≥ 11
and x ≥ 0, y ≥ 0
To solve the LPP we convert the inequations into equations as follows :
3x + 4y = 8
5x + 2y= 11;
x = 0 = y
For region 3x + 4y ≥ 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).

For region 5x + 2y ≥ 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.
Also x ≥ 0, y ≥ 0 represent the first quatrant.
Both lines intersects at B(2, \(\frac{1}{2}\))
The 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
The values of the objective function at these corner points are given in the following table :

Corner Point	Value of objective function Z = 60x + 80y
A(\(\frac{8}{3}\), 0)	Z = 160
B(2, \(\frac{1}{2}\))	Z = 160
C(0, \(\frac{11}{2}\))	Z = 440

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

Question 22(Old).
A 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.
To 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 ?
Answer:
Let the cottage industry manufacture x pedestal lamps and y wooden shades.
Therefore, x ≥ 0 and y ≥ 0
The given infromation can be tabulated as follows :

The profit on a lamp is ? 5 and on the shades is ? 3. Therefore, the constraints are
2x + y ≤ 12
3x + 2y≤ 20
Total profit, Z = 5x + 3y, we have to maximise Z.
The mathematical formulation of the given problem is given below :
Maximize Z = 5x + 3y
subject to the constraints,
2x + y ≤ 12
3x + 2y ≤ 20
x, y ≥ 0
Region 2x + y ≤ 12 ; The line 2x+y= 12 meet coordinate axes at A(6,0) and B₁(0,12). region containing (0,0) gives the solution set of 2x + y ≤ 12 since (0,0) lies on 2x + y ≤ 12.
Region 3x + 2y ≤ 20 : The line 3x + 2y = 20 meet coordinate axes at A₂(\(\frac{20}{3}\),0) and C(0, 10). The region containing (0,0) be the solution set of given inequation.
The region x ≥ 0, y ≥ 0 represents the first quadrant of xy plane.
Both lines intersects at B(4, 4).
The 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).
The value of Z at these corner points are as follows:

Corner Point	Z = 5x + 3y
O(0, 0)	0
A1(6, 0)	30
P(4, 4)	32 → Maximum
B2(0, 10)	30

The maximum value of Z is 32 at (4,4).
Thus, 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.

Question 23.
A 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 ₹ 5/kg and B costs ₹ 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)
Answer:
Let x kg of fertiliser A and y kg of fertiliser B should be required to minimise the cost.
Given information can be put in tabular form as below :

Fertilizers	A	B	Min. Requirements
Nitrogen	0.1x	0.05y	7 kg
Phosphoric	0.06x	0.1y	7 kg
Min. Cost	5x	8y

Let Z be the total cost ∴ Z = 5x + 8y
So our aim is to minimize Z
Then the mathematical modelling of the LPP is given as follows :
Minimize Z = 5x + 8y
Subject to constraints;
0.1x + 0.05y ≥ 7
0.06x + 0.1y ≥ 7
and x, y ≥ 0
For region 0.1x + 0.05y ≥ 7 ;
The 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.
Thus the region not containing (0, 0) gives the solution set of given region.
Region 0.06x + 0.01y ≥ 7 ;
The line 0.06x + 0.1y = 7 meet coordinate axes at C(\(\frac{350}{3}\), 0) and
D(0, 70). Since (0, 0) does not satisfies the set of given region.
Thus the region not containing (0, 0) gives the solution set of given region.
Further x ≥ 0, y ≥ 0 represents the first quadrant of XOY plane.

Further 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.

Corner Point	Z = 5x + 8y
C(\(\frac{350}{3}\), 0)	Z = 5 × \(\frac{350}{3}\) + 8 × 0 = \(\frac{1750}{3}\)
P(50, 40)	Z = 5 × 50 + 8 × 40 = 570
B(0, 140)	Z = 5 × 0 + 8 × 140 = 1120

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.
Thus smallest value be the minimum value.
Z_min = 500 at x = 50 and y = 40
Hence 50 kg of fertiliser A and 40 kg of fertiliser B should be bought to get the minimum cost of ₹ 570.

Question 24.
A 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 ₹ 12 and ₹ 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)
Answer:
Let 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.
Then the mathematical modelling of the LPP is given as follows :
Maximize Z = 12x+ 16y
Subject to constraints ;
x + y ≤ 1200
\(\frac{1}{2}\)x – y ≥ 0
x – 3y ≤ 600
x ≥ 0, y ≥ 0
To solve the LPP we convert the given inequations into equations x + y = 1200
x + y = 1200
\(\frac{1}{2}\)x – y = 0
x – 3y = 600
Region x + y ≤ 1200 ; The lin ex+y= 1200 meets coordinate axes at A₁ (1200, 0) and B₁(0,1200). Since (0, 0) satisfies the x + y ≤ 1200. So the region containing (0, 0) gives the solution set of given inequation.
Region \(\frac{1}{2}\)x – y > 0; The line \(\frac{x}{2}\) – y = 0 passes through origin.
Region x – 3y ≤ 600 : The line x -3y = 600 meets the coordinate axes at A₂ (600,0) and B₂ (0, – 200). Since (0, 0) satisfies the inequations.
Hence the region containing (0, 0) gives the solution set of inequation.
Also, x ≥ 0, y ≥ 0 represent the first quadrant of xoy plane.
The lines x + y = 1200 and \(\frac{x}{2}\) – y = 0 intersect at P (800, 400).
The lines x + y = 1200 and x – 3y = 600 intersect at Q (1050, 150)
The feasible region of the LPP is shaded in graph.
The shaded area OA2QP gives the feasible region
Thus, the coordinates of the vertices of (Comer – points) of shaded feasible region OA₂PQ are O (0, 0) ; A₂ (600, 0), Q (1050, 145) and P (800, 400).

The values of the objective function at corner points are given in the following table:

Corner Point (x, y)	Value of objective function Z = 12x + 16y
A2 (600, 0)	Z = 12 × 600 + 16 × 0 = 7200
Q (1050, 145)	Z= 12 × 1050 + 16 × 145 = 14920
P(800, 400)	Z = 12 × 800 + 16 × 400 = 16000
O (0, 0)	Z = 0

Clearly Z_max = 16000 at x = 800 and y = 400
Thus, the toy company should manufacture 800 dolls of type A and 400 tolls of type B to earn maximum profit.
Thus, the required maximum profit that can be earned by company is ₹ 16,000.

Question 24(Old).
A 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 ₹ 20 and on one doll of type II is ₹ 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)
Answer:
Let 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 ₹ 20 and ₹ 30 respectively. So profit gained by comnpany on selling x units of doll I and y units of doll II be 20x + 30y.
Let Z be the total profit of company. ∴ Z = 20x + 30y we have to maximize Z.
Since 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 ;
1.5x + 3y ≤ 42 => x + 2y ≤ 28
Also 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
∴ 3x + y ≤ 24
The mathematical formulation of given LPP is as under ;
Max Z = 20x + 30y
Subject to constraints ;
x + 2y ≤ 28
3x + y ≤ 24
x, y ≥ 0 [since, dolls of I and II can’t be negative]
To solve LPP, we convert inequations into eqn’s :
For region x + 2y ≤ 28 ; The linex + 2y = 28 meets coordinate axis at A (28,0) and B(0, 14). Since (0, 0) lies on x + 2y ≤ 28. So the region containing (0, 0) gives the soln. set of given inequation. For region 3x + y ≤ 24 ; The line 3x+y = 24 meets coordinate axis at C (8, 0) and D (0, 24). Since (0, 0) lies on 3x + y ≤ 24.
So region containing (0, 0) gives the soln. set of 3x + y ≤ 24.
Region x ≥ 0, y ≥ 0 represents the first quadrant of XOY plane.
Both lines intersects at P (4, 12).
Thus, 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).

We evaluate Z at these corner points

Corner point	Z = 20x + 30y
O (0, 0)	0
C (8, 0)	160
P(4, 12)	80 + 360
B (0, 14)	0 + 30 × 14 =  420

Z_max = 440 at P (4, 12) i.e. at A = 4 and y = 12
∴ required no. of doll of type 1 = 4; required no. of doll II = 12 and Maximum profit = ₹ 440.
Since 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.

Question 25.
The feasible region for an L.P.P. is shown in the adjoining figure.

(i) Write the constraints for the L.P.P.
(ii) Write the coordinates of the points A, B and C.
(iii) Find the maximum value of the objective function Z = 30x + 50y.
Answer:
(i) The eqn. of AD is
\(\frac{x}{24}+\frac{y}{24}\) = 1 i.e. x + y = 24
The eqn. of EC is
\(\frac{x}{32}+\frac{y}{16}\) = 1 ⇒ x + 2y = 32
As, (0, 0) lies in x + y ≤ 24 and (0, 0) also satisfies x + 2y ≤ 32.
Thus associated constraints for given L.P.P. are ; x + y ≤ 20 ; x + 2y ≤ 32, x ≥ 0, y ≥ 0
(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).
(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).

Corner points	Z = 30x + 50y
0 (0, 0)	Z = 30 × 0 + 50 × 0 = 0
A (24,0)	Z = 30 × 24 + 50 × 0 = 720
B (16, 8)	Z = 30 × 16 + 50 × 8 = 480 + 400 = 880
C (0, 16)	Z = 30 × 0 + 50 × 16 = 800

Clearly Z_max = 880 at B (16, 8)
i.e. at x = 16 and y = 8

Question 26.
The feasible region for an L.P.P. is shown in the adjoining figure. The line CB is parallel to OA. Answer the following :

(i) Write the constraints for the L.P.P.
(ii) Write coordinates of the point B.
(iii) Find the minimum value of the objective function Z = 3x-4y
Answer:
(i) The eqn. of line CB be given by
y – 4 = \(\frac{16-4}{6-0}\)(x – o)
⇒ y – 4 = 2x ⇒ y = 2x + 4
The eqn. of line OA be given by 12-0
y – 0 = \(\frac{12-0}{6-0}\)(x – 0)
⇒ y = 2x
Clearly (0, 0) satisfies y – 2x ≤ 4 and (0, 0) satisfies y ≥ 2x.
Also (0, 0) satisfies x ≤ 6.
Thus the associated constraints for given L.P.P is as under : y – 2x ≤ 4; y ≥ 2x, x ≤ 6; x ≥ 0, y ≥ 0

(ii) Point B be the point of intersection of lines x = 6 and y – 2x = 4 ,Coordinates of B are (6, 16).
(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).

Corner points	Z = 3x – 4y
0 (0, 0)	Z = 3 × 0 – 4 × 0 = 0
C (0, 4)	Z = 3 × 0 – 4 × 4 = – 16
B (6, 16)	Z = 3 × 6 – 4 × 16 = – 46
A (6,12)	Z = 3 × 6 – 4 × 12 = -30

Question 26(Old).
Two tailors P and Q earn ₹ 150 and ₹ 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.
Answer:
Suppose tailor A and B work for x and y days respectively to minimise the cost.
Since tall or A and B earn ₹ 150 and ₹ 200 respectively So, tailor A and B earning for x and y days be ₹ 150 x and 200y respectively, let Z denote maximum profit that gives minimum labour cost Then Z = 150x + 200y
Since, 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
6x + 10y ≥ 60
3x + 5y ≥ 30
Since, 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
4x + 4y ≥ 32
x + y ≥ 8
Hence, The required mathematical formulation of LPP is, as given below Min. Z= 150.x+ 200y
Subject to constraints,
3x + 5y ≥ 30
x + y ≥ 8
x, y ≥ 0 [Since x and y not be less than zero]
Region 3x + 5y ≥ 30: line 3x + 5y = 30 meets coordinate axes at A₁(10,0) and B₂(0,6) respectively. Region not containing origin represents the solution set of inequation 3x + 5y > 30, as (0,0) does not satisfy 3x + 5y ≥ 30.
Region x + y ≥ 8: line x + y = 8 meets axes at C₁(8,0) and D₁(0,8) respectively. Region not containing origin represents the solution set of inequation x + y ≥ 8 as (0,0) does not satisfy x + y ≥ 8.

Region x, y ≥ 0 : it represent the first quadrant in the xy plane,
both given lines 3x + 5y = 30 and x + y = 8 intersects at E₁(5, 3).
Unbounded shaded region A₁E₁D₁ represents the feasible region with corner points A₁(10,0), E₁(5, 3) and D₁(0.8).

Corner point	Z = 150x + 200y
A1(0, 0)	150(10) + 200(0) = 1500
E1 (5, 3)	150(5) + 200(3) = 1350
D1(0, 8)	150(0) + 200(8) = 1600

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,
Zmin = 1350, at x = 5, y = 3
Thus, Tailor A should work for 5 days and B should work for 3 days.

Question 27.
An 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 :

Assuming that the transportation cost per km is ₹ 1 per litre, how should the delivery be scheduled in order that the transportation cost is minimum ? What is the minimum cost? (NCERT)
Answer:
Let 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.
Since 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.
Similarly, (3000 -y) Land 3500 -(7000- x-y) = (x +y- 3500) L will be transported from depot B to petrol pump E and F respectively.
The given problem can be represented diagramatically given as follows.

x ≥ 0, y ≥ 0 and (7000 – x – y) ≥ 0 x ≥ 0, y ≥ 0 and x + y ≤ 7000 i.e. 4500 – x ≥ 0, 3000 – y ≥ 0, and x + y – 3500 ≥ 0 Thus, x < 4500, y ≤ 3000, and x + y ≥ 3500
Given cost of transporting 10 L of petrol = Re 1
Cost of transporting 1 L of petrol = ₹ \(\frac{1}{10}\)
Therefore, total transportation cost is given by,
Z = \(\frac{7}{10}\) 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)
i.e. Z = 30x + 0.1y + 3950
The given problem can be formulated as follows :
Minimize Z = 0.3x + 0.1y + 3950
subject to the constraints,
x +y ≤ 7000
x ≤ 4500
y ≤ 3000
x + y ≥ 3500
x, y ≥ 0
The feasible region determined by the constraints is as follows.
For region x + y ≤ 7000 ; The line x + y = 7000 meets coordinate axes at A₁(7000, 0) and B₁(0, 7000). Region containing (0,0) represents the solution set of inequation.
For region x ≤ 4500; The line x = 4500 is parallel to y-axis meeting x-axis at C,(4500,0).
Since (0. 0) lies on x ≤ 4500.
Thus region containing (0,0) represents the solution set of inequation x ≤ 4500.
Forregiony ≤ 3000 ; The line y ≤ 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 ≤ 3000.
Since (0, 0) lies on y ≤ 3000.
For region x + y ≥ 3500 : The line x + y = 3500 meets coordinate axes at E₁ (3500, 0) and F₁ (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.
The line x = 4500 meets line x + y = 7000 at I₁ (4500, 2500) and the line y = 3000 meets the line x + y = 7000 at H,(4000, 3000)
The line y = 3000 meet x + y = 3500 intersects at G, (500, 3000)
The shaded area E₁C₁I₁H₁G₁ gives the feasible region.
The corner points of the feasible region are E₁(3500,0), C₁(4500,0), I₁(4500, 2500), H₁,(4000, 3000) and G₁(500, 3000).

The value of Z at these corner points are as follows :

Corner point	Z = 0.3x + 0.1y + 3950
E1(3500,0)	5000
C1(4500,0)	5300
I1 (4500, 2500)	5550
H1 (4000, 3000)	5450
G1 (500,3000)	4400 → Minimum

The minimum value of Z is 4400 at (500, 3000).
Thus, 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 ₹ 4400.

Question 29(Old).
David wants to invest atmost ₹ 12000 in Bonds A and B. According to the rule, he has to invest atleast ₹ 2000 in Bond A and atleast ₹ 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.
Answer:
Let David invest ₹ x in bond A and ₹ y in bond B.
It is given that, rates of interest on bonds A and B respectively are 8% and 10% per annum.
So Total interest gained by david = \(\frac{8 x}{100}+\frac{10 y}{100}\)
Let Z be the total interest gained by David
Then Z = \(\frac{8 x}{100}+\frac{10 y}{100}\) and we want to maximise Z.
Since David wants to invest atmost ₹ 12000 in bonds A and B. Then investment constraint is given by x + y ≤ 12000
Since, David has to invest atleast ₹ 2000 in bond A and atleast ? 4000 in bond B. x ≥ 2000 and y ≥ 4000
Thus the mathematical modelling of given LPP is as under :
Max Z = \(\frac{8 x}{100}+\frac{10 y}{100}\)
Subject to constraints ; x + y ≥ 12000 ;
x ≥ 2000 ; y ≥ 4000
x, y ≥ 0 [since investment can’t be negative]
To solve LPP, we convert inequations into equations :
x + y = 12000 ; x = 2000 ;y = 4000
For region x + y ≤ 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.
For region x ≥ 2000 ; The line x = 2000 meets x-axis at C (2000, 0) and (0, 0) does not lies on x ≥ 2000 and region not containing (0, 0) represents the soln. set of x > 2000.
For region y ≥ 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 ≥ 4000.
x ≥ 0, y ≥ 0 represents the first quadrant of XOY plane.
The lines x = 2000 and x + y = 12000 meets at P (2000, 10000)
The lines y = 4000 and x + y = 12000 at Q (3000, 4000)
both lines x = 2000 and y = 4000 intersects at R (2000, 4000)
We evaluate Z at these corner points.

Corner point	Z = \( \frac{8 x}{100}+\frac{10 x}{100} \)
P(2000, 1000)	\( \frac{8}{100} \) x 2000 + \( \frac{10}{100} \) x 1000 = 1160 (Max)
Q(8000, 400)	\( \frac{8}{100} \) x 800 + \( \frac{10}{100} \) x 400 = 1040
R(2000, 4000)	\( \frac{8}{100} \) x 200 + \( \frac{10}{100} \) x 4000 = 560

∴ Z_max = 1160 at x = 2000 ; y = 10,000
Hence, Required investment in bond A = ? 2000 Required investment in bond B = ₹ 10000 and Max. Interest = ₹ 1160

Question 30(Old).
A farmer mixes two brands P and Q of cattle feed. Brand P costing ₹ 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 ₹ 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)
Answer:
Let 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.
Since costing of brand P and brand Q per bag be ₹ 250 and ₹ 200 respectively.
So cost of x units of bags of brand P andy meets of bags of bag Q be 250x and 200y.
Total cost of cattle feed be 250x + 200y
Let Z be the total cost of cattle feed then Z = 250x + 200y.
Since 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 ⇒ 2x + y ≥ 12
Similarly brand P and Q contains 25 units and 11.25 units of element B and minimum requirement 4.5 units of element B be
Second Constraints is given by 2.5x + 11.25y
Similarly 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 :
Max Z = 250x + 200y Subject to constraints ;
2 x + y ≥ 12 1 Ox + 45y ≥ 18
2x + 3y ≥ 24 and x, y ≥ 0 [since bags of cattle feed can’t be negative]

To solve LPP, we convert inequations into equations :
2x + y = 12 ;
10x + 45y = 18 ;
2x + 3y = 24 ;
x = y = 0
For region 2x + y ≥ 12 ;
The line 2x+ y= 12 meets coordinate axis at A (6, 0) and B (0, 12)
Since (0, 0) does not lies on given region.
Hence region not containing (0, 0) gives the soln. set of given inequation.
For region 10x + 45y ≥ 18 ;
The 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.
For region 2x + 3y ≥ 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 ≥ 24. x ≥ 0, y ≥ 0 represent the first quadrant of XOY plane.

The lines 2x + y = 12 and 10x + 45y = 18 intersects at P\(\left(\frac{261}{40},-\frac{21}{20}\right)\)
The lines 10x + 45y = 18 and 2x + 3y = 24 intersects at Q\(\left(\frac{171}{10},-\frac{34}{10}\right)\)
The lines 2x + y = 12 and 2x + 3y = 24 intersects at R (3, 6).
The shaded region ERB be the feasible region and is unbounded.
We evaluate Z at these points E (12, 0), R (3, 5) and B (0, 12).

Corner point	Z = 250x + 200y
E(12, 0)	250 × 12 = 3000
R(3, 6)	250 × 3 + 200 × 6 = 1950 (Min)
B(0, 12)	250 × 0 + 200 × 12 = 2400

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.
∴ Z_min = 1950 at x = 3 and y = 6
required no. of bags of brand P = 3
and required no. of bags of brand Q = 6
and minimise cost = ₹ 1950