Question

Evaluate maximum and minimum value, where Z = 3x + 9y Subject to the constraints x + 3y ≤ 60, x + y ≥ 10, x – y ≥ 0 and x ≥ 0, y ≥ 0

Answer 1

Converting the given in equations into equations

x + 3y = 60 …..(1) 

x + y = 10 …..(2)

 x = 0, y = 0 ….(3) 

Region represented by x + 3y ≤ 60: 

The line x + 3y = 60 meets the coordinate axis at A(60,0) and B(0, 20). x + 3y = 60

x	60	0
y	0	20

A(60, 0); B(0, 20)

Join the points A to B to obtain the line.

Clearly (0,0) satisfies the in equation x + 3y ≤ 60.

So the region containing the origin, represents the solution set of the in equation.

Region represented by x + y ≥ 10 : 

The line x + y = 10 meets the coordinate axis at point C(10,0) and D(0, 10).
x + y = 10

x	10	0
y	0	10

C(10, 0);D(0, 10)

Join point C to D to obtain the line.

Clearly (0, 0) does not satisfy the in equation x + y ≥ 10.

So the region opposite to origin, represents the solution set of the in equation.

Region represented by x – y ≥ 0 : 

The line x – y = 0 meets the coordinate axis at O(0, 0), E(5, 5).

x	0	5
y	0	5

O(0, 0); E(5, 5)
Join point O to E to obtain the line. 

Clearly (0, 0) satisfies the x – y ≥ 0. 

So the region cointaining origin represents the solution of in equation.

Region represented by x ≥ 0, y ≥ 0 : 

Since every point in the first quadrant satisfies these in equations. 

So the first qudrant is the region represented by the in equation x ≥ 0 and y ≥ 0.

The shaded region BDEF represents the solution region of the above in equations. This region is the feasible region of the given L.P.P.

The coordinates of the comer points of the shaded feasible region are (0, 10), (5, 5), (15, 15) and (0, 20). 

The point C(15, 15) is the point of intersection of lines x =y and x + 3y=60 and point E is the intersection point of lines x + y = 10 and x = y.

The values of the objective function on these points are given in the following table:

Point	x-coordinate	y-coordinate	Objective function Z = 3x + 9y
D	0	10	ZP = 3(0) + 9(10) = 90
E	5	5	ZE = 3(5) + 9(5) = 60
F	15	15	ZF = 3(15) + 9(15) = 180
B	0	20	ZH = 3(0) + 9(20) = 180

Clearly value of objective function Z is minimum at point E(5,5) where x = 5 and y = 5 and Z= 60 and the Z is maximum at two points C( 15, 15) and D(0, 20) where Z = 180.

Hence the Minimum value of Z = 60 and Maximum value of Z = 180.