Question

Solve the following linear programming problem by the graphical method.

Maximize Z = 400 X1 + 200 x 2

Subject to : X1 + 8X2 <= 24

X1 + 2X2 <= 12

X1 >= 0 , X2 >= 0

You will need to graph each of the constraints to answer the following questions. You can draw a rough graph.

a) State the coordinates of the point where the constraints interact.

b) Define in words the region of feasible solutions.

c) Lost all of the vertices or corner points on the region of feasible solutions.

d) Find the solution which will maximize profit (Z) and give the numerical value for the maximum profit.

Answer 1

MAXIMIZE Z = 400 X1 + 200 X2

Subject to:

X1 + 8X2 <= 24

X1 + 2X2 <= 12

X1 >= 0 , X2 >= 0

Putting in Graph:

In x-axis we have X1. In y-axis we have X2.

the blue line is for constraint X1+8X2 <=24

the orange line is for constraint X1+2X2<= 12.

Question – a:

Points where the constraints meet:

We have 4 constraints.

(0,0), (12,0), (0,3), (8,2), (25,0), (0,6)

Question – b:

The region of feasible solution:

The region of feasible solution is enclosed by the area with the corner points (0, 0), (12, 0), (0, 3), (8, 2)

Hence the area is bound by the two axes, and the two constraint lines combined.

This is because the constraints X1 + 8X2 <= 24 and X1 + 2X2 <= 12 encloses the area as lower the line area and X1 >= 0 and X2 >= 0 mentioning the area be in Quadrant 1.

Question – c:

List all of the vertices or corner points on the region of feasible solutions: (0,0), (12,0), (0,3), (8,2)

Question – d:

Putting the four points in the Maximization line:

MAX Z = 400 X1 + 200 X2

(0,0) will give 0

(12,0) will give 12*400 = 4800

(0,3) will give 3*200 = 6

(8,2) will give 400*8 + 2*200 = 3200 + 400 = 3600

The maximum value is for (12,0)

Hence the solution is: x1 = 12; x2 = 0 Maximum value is 4800

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