Answer to Question #190701 in Operations Research for Raa

Question

To solve the ILLP given below by the graphical method :

Maximum Z = 95x₁ + 100x₂

Subject to the constraints

5x₁ + 2x₂ <= 20

x₁ >= 3

x₂ <= 5

x₁ , x₂ are non-negative integers.

Answer 1

$Z=95X_1+100X_2$ (Maximize)

The condition given:

$5X_1+2X_2\leq 20$

$X_1\geq3$

$X_2\leq5$

In the equation $5X_1+2X_2=20$

when we put $X_1=0$,then $X_2=10$

When we put $X_2=0$,then $X_1=4$

Now when we draw a graph of the line and the given points and shade accordingly then in the feasible region we got three points

that is

A(3,0)

B(4,0)

C(3,2.5)

Now value of Z at A=

$Z_A=95\times3+0=285$

Value of Z at B=

$Z_B=95\times4+0=380$

Value of Z at C=

$Z_C=95\times3+100\times2.5=535$

So the maximum value is achieved at the point C and that value is 535

So we can say that the Maximum of Z is 535 and it occurred at the point (3,2.5)