OP Malhotra Class 12 Maths Solutions Chapter 28 Linear Programming Ex 28(b)

Students can track their progress and improvement through regular use of S Chand Class 12 Maths Solutions Chapter 28 Linear Programming Ex 28(b).

S Chand Class 12 ICSE Maths Solutions Chapter 28 Linear Programming Ex 28(b)

Question 1.
Find the maximum value of each expression on region R
(i) 3x + 4y
(ii) 2y – x

Solution:
(i) The corner points of region R are given as under:

Corner points	Z = 3x + 4y
O (0,0)	3 × 0 + 4 × 0 = 0
A (0,4)	3 × 0 + 4 × 4 = 0
B (3,4)	3 × 3 + 4 × 4 = 25 (Max)
C (5,2)	3 × 5 + 4 × 2 = 23
D (6,0)	3 × 6 + 4 × 0 = 18

Thus Z_max = 25 at B (3, 4) i.e. x = 3 and y = 4

(ii)

Corner points	Z = 2y – x
O (0,0)	0
A (0,4)	2 × 4 – 0 = 8 (Max)
B (3,4)	2 × 4 – 3 = 5
C (5,2)	2 × 2 – 5 = -1
D (6,0)	2 × 0 – 6 = -6

∴ Z_max = 8 at A (0, 4) i.e. x = 0 and y = 4

Question 2.
Find the minimum value of each expression on region S shown in

(i) x + y
(ii) 5x + 4y
Solution:

Corner points	Z = x + y
E (0,10)	0 + 10 = 10
F (5,2)	5 + 2 = 7 (Min)
G (7,0)	7 + 0 = 7 (Min)

∴ Z_min = 7 at F (5, 2) and G (7, 0)
i.e. at the line segment joining F and G.

(ii)

Corner points	Z = 5x + 4y
E (0,10)	5 × 0 + 4 × 10 = 40
F (5,2)	5 × 5 + 4 × 2 = 33(Min)
G (7; 0)	5 × 7 + 4 × 0 = 35

∴ Z_min = 33 at F (5, 2) i.e. at x = 5 and y = 2

Question 3.
(i) Evaluate the linear expresion 3x + 2y at each corner point of the convex polygonal region shown in the graph below :

(ii) What is the minimum value of 3x + 2y over the region shown in the graph above ?
Solution:

Corner points	Z = 3x + 2y
(8,0)	3 × 8 + 2 × 0 = 24
(4,1)	3 × 4 + 2 × 1 = 14
(2,3)	3 × 2 + 2 × 3 = 12(Min)
(0,7)	3 × 0 + 2 × 7 = 14

(ii) ∴ Z_min = 12 at x = 2 and y = 3

Question 4.
Refer to the following system of inequalities x ≥ 1, x ≥ 2, 3y + x ≥ 9, y + 2x ≥ 8
(i) Graph the solution set of the system.
(ii) Determine the corner points of the region that represents the solution set.
(iii) Find the minimum value of each of the following linear expresions over the region graphed in (i) by evaluating the expression at each of the corner points.
(a) x + 5y
(b) 5x +y
(c) x+y
Solution:
First of all, converting the given inequalities into equation
y= 1 ; x = 2 ; 3y + x – 9 and y + 2x = 8
For region y ≥ 1 ; The liney = 1 be a line parallel to x-axis and pass through A (0, 1). Since (0, 0) does not satisfies y ≥ 1. Thus region not containing (0, 0) gives the soln. set of region y ≥ 1.
For region x ≥ 2; The line x = 2 parallel to y-axis and passing through the point B (2,0). Since (0,0) does not satisfies x ≥ 2. Thus the region not containing (0, 0) gives the soln. set of region x ≥ 2.

For region 3y + x ≥ 9 ; The line 3y + x = 9 meets coordinate axis at C (9,0) and D (0,3). Since (0, 0) does not satisfies 3y + x ≥ 9. Thus region not containing (0,0) gives the soln. set of given region.
For region y + 2x ≥ 8 ; The line y= 2x + 8 meets coordinate axes at E (4, 0) and F (0, 8). Since (0, 0) does not satisfies y + 2x ≥ 8. Thus region not containing (0, 0) gives the soln. set of given region. ‘

The line y = 1 intersects 3y + x = 9 at P (6, 1).
Both lines 3y + x = 9 and y + 2x = 8 intersects at Q (3, 2).
The lines x = 2 and y + 2x = 8 intersects at R (2, 4).
Here the shaded feasible region is unbounded and its comer points are
P (6, 1); Q (3, 2) and R (2, 4).

(iii)

(a) Corner points	Z = x + 5y
P (6,1)	6 + 5 × 1 = 11(Min)
Q (3,2)	3 + 5 × 2 = 13
R (2,4)	2 + 5 × 4 = 22

∴ Z_min = 11

(b) Corner points	Z = 5x + y
P (6, 1)	5 × 6 + 1 = 31
Q (3, 2)	5 × 3 + 2 = 17
R (2, 4)	5 × 2 + 4 = 14 (Min)

∴ Z_min = 14

(c) Corner point	Z = x + y
P (6,1)	6 + 1 = 7
Q (3,2)	3 + 2= 5(Min)
R (2,4)	2 + 4 = 6

Question 5.
Refer to the following system of inequalities.
x ≥ 1, y ≤ 6, y ≥ – x + 4, 2y ≥ x – 1 ,3y ≤ -x + 21
(i) Graph the solution set of the system and label the corner points with their coordinates,
(ii) Find the minimum value of the expression x + 5y over the region graphed in (i).
(iii) Find the maximum value of the expression 4x – 3y over the region graphed in (i).
Solution:
Converting the given inequalities into equations :
x = 1 ; y = 6; y = -x + 4; 2y = x – 1 ; 3y = – x + 21
For region x ≥ 1 : The line x = 1 is parallel to y-axis and passes through the point A (1, 0) and (0,0) does not lies on x ≥ 1. Thus region not containing (0, 0) gives the soln. set of given region x ≥ 1. For region y ≤ 6 ; The line y = 6 is parallel to x-axis and passes through the point B (0, 6). Also (0, 0) satisfies the region y ≤ 6. Thus the region containing the (0, 0) gives the soln. set of given region y ≤ 6.

For region y ≥ – x + 4 ; The line y = – x + 4 meets coordinate axes at C (4, 0) and D (0, 4). Since (0,0) does not satisfies y ≥ – x + 4. Thus region not containing (0,0) gives the soln. set of given region.

For region 2y ≥ x – 1 ; The line 2y = x – 1 meets coordinate axes at E (1,0) and F \(\left(0,-\frac{1}{2}\right)\) since (0, 0) satisfies the region 2y ≥ x – 1. Thus the region containing the origin gives the soln. set of given region.

For region 3y ≤ – x + 21; The line 3y = – x + 21 meets coordinate axes at G (21, 0) and H(0, 7). Since (0, 0) satisfies the region 3y ≤ – x + 21. The region containing (0,0) gives the soln. set of given region.

The lines x + y = 4
and 2y = x- 1 intersect at P (3, 1)
The lines 2y = x – 1
and 3y = – x + 21 intersects at Q (9, 4).
The lines x = 1
and x + y = 4 intersects at R (1, 3).
The lines 3y = – x + 21 meets the line y = 6 at T (3, 6).
The lines x = 1 and y = 6 intersects at S (1, 6).
Thus the shaded feasible region PQ ⊥ SR be the bounded region and its corner points are P (3, 1) ; Q (9, 4) ; T (3, 6) ; S (1, 6) and R (1, 3).
(ii)

Corner points	Z = x + 5y
P (3,1)	3 + 5 × 1 = 8(Min)
Q (9,4)	9+5 × 4 = 29
T (3,6)	3+5 × 6 = 33
S (1,6)	1+5 × 6 = 31
R (1,3)	1 + 5 × 3 = 16

∴ Z_min = 8
(iii)

Corner points	Z = 4x – 3 y
P (3,1)	4 × 3 – 3 × 1 = 9
Q (9,4)	4 × 9 – 3 × 4 = 24(Max)
T (3,6)	4 × 3 – 3 × 6 = -6
S (1,6)	4 × 1 – 3  × 6 = -14
R (1,3)	4 × 1 – 3  × 3 = -5

Question 6.
Minimize Z = 20x + 10y subject to the constraints
x + 2y ≤ 40
3x + y ≥ 30
4x + 3y ≥ 60
x ≥ 0 ; y ≥ 0
Solution:
First of all, we convert all given inequalities into eqn’s
i.e. x + 2y = 40 ; 3x + y = 30 and 4x + 3y = 60 ; x = 0 = y
For region x + 2y ≤ 40 ; The line x + 2y = 40 meets x-axis at A (40, 0) and y-axis at B (0, 20). Clearly (0, 0) satisfies the given region (∵ 0 + 2 × 0 ≤ 40, which is true)
Thus, region containing (0, 0) gives the soln. set of given region.

For region 3x + y ≥ 30 ; The line 3x + y = 30 meets coordinate axes at C (10, 0) and D (0, 30). Clearly (0, 0) does not satisfies the region 3x + y ≥ 30. Thus, the region not containing (0, 0) gives the solution set of given region.
For region 4x + 3y ≥ 60 ; The line 4x + 3y = 60 meets coordinate axes at E (15, 0) and F (0, 20). Clearly (0, 0) does not satisfies the given region. Hence the region not containing (0,0) gives the soln. set of given region.
For region x ≥ 0 ; y ≥ 0 represents the first quadrant of XOY plane.
The lines 3x + y = 30 and 4x + 3y = 60 intersects at P (6, 12).
The lines 3x + y – 30 and x + 2y = 40 intersects at Q (4, 18).
Thus the shaded region EPQA be the bounded feasible region and its corner points are E (15, 0); A (40, 0); P (6, 12) and Q (4, 18).
Now we evaluate Z at these comer points.

Corner points	Z = 20x + 10y
E (15,0)	20 × 15 + 10 × 0 = 300
A (40,0)	20 × 40 + 10 × 0 = 800
P (6,12)	20 × 6 + 10 × 12 = 240
Q (4,18)	20 × 4 + 10 × 18 = 260

∴ Z_min = 240 at P (6, 12) i.e. at x = 6 and y = 12

Question 7.
Minimize Z = 3x + 4y subject to constraints
x + y ≥ 3
2x + y ≥ 4
x, y ≥ 0
Solution:
Converting the given constraints into equations x + y = 3 ;2x + y = 4; x = y = 0
The region x + y ≥ 3 ; The line x + y = 3 meets coordinate axes at A (3, 0) and B (0, 3). Clearly (0,0) does not lies on region x + y ≥ 3. Thus region not containing (0,0) gives the solution set of given region.
For region 2x + y ≥ 4 : The line 2x + y = 4 meets coordinate axes at C (2, 0) and D (0, 4). Clearly (0, 0) does not satisfies given region. Thus the region not containing (0, 0) gives the solution set of given region.
Clearly x ≥ 0 ; y ≥ 0 represents the first quadrant of XOY plane.
Here both lines x + y = 3
and 2x + y = 4 intersects P (1, 2).
Here the shaded region APD be the feasible region and it is unbounded.
The comer points of feasible region are A (3, 0); P (1, 2) and D (0, 4).

Corner points	Z = 3x + 4y
A (3,0)	3 × 3 + 4 × 0 = 9
P (1,2)	3 × 1 + 4 × 2 = 11
D (0,4)	3 × 0 + 4 × 4 = 16

Here smallest value of Z be 9 at A (3, 0).
Since the region is unbounded. So we check whether this smallest value is minimum or not. For this, we check open half plane 3x + 4y ≤ 9 have common points in feasible region or not. Clearly open half plane have no common points in feasible region.
∴ Z_min = 9 at x = 3 and y = 0.

Question 8.
Maximize Z = 6x₁ + 11x₂ subject to the constraints
2x₁ + x₂ ≤ 104
X₁ + 2x₂ ≤ 76
X₁, x₂ ≥ 0
Solution:
Converting the given inequalities into equations ;
2x₁ + x₂ = 104 and X₁ + 2x₂ = 76 ; x₁ = x₂ = 0
For region 2x₁ + x₂ ≤ 104 ; The line x₁ + x₂ = 104 meets coordinate axes at A (52, 0) and B (0, 104). Clearly (0,0) satisfies the given region. So region containing (0, 0) gives the solution set of given region.

For region x₁ + 2x₂ ≤ 76; The line x₁ + 2x₂ = 76 meets coordinates axes at C (76, 0) and D(0, 38). Clearly (0, 0) satisfies the given region. So region containing (0, 0) gives the solution set of given region.

Also, x₁ ≥ 0 ; x₂ ≥ 0 represents the first quadrant of first quadrant.
Further both given lines 2x₁ + x₂ = 104
and x₁ + 2x₂ = 76 intersects at P (44, 16).
Thus the shaded region OAPD be the bounded feasible region and its comer points are
O (0, 0); A (52, 0); P (44, 16) and D (0, 38).

Corner points	Z = 6x1 + 11 x2
O (0,0)	0
A (52,0)	6 × 52 + 11 × 0 = 312
P (44,16)	6 × 44 + 11 × 16 = 440
D (0,38)	6 × 0 + 11 × 38 = 418