Question

Maximize:4x₁+12x₂

subject to:

3x₁+x₂<=180

x₁+2x₂<=100

-2x₁+2x₂<=40

X₁>=0

x2>=0

Find the optimal solution for the above model

Answer 1

To find the optimal solution we have to draw the graph of the graph of the given constraints making all the given inequation as equation such as :

$L_1:3x_1+x_2=180$

$L_2:x_1+2_2=100$

$L_3:-2x_1+2x_2=40$

Now we draw the graph as follows:

Feasible solution of the given L.P.P was shown in the shaded region $OABCDO$ .

Since the shaded region is bounded, each corner points gives the optimal solution of the given L.P.P.

Solving $L_1$ and $L_2$ we get the corner point $B(52,24)$

Solving $L_1$ and $L_3$ we get the corner point $C(20,40)$

And also $A(60,0)$ , $D(0,20)$ and $O(0,0)$ are the corner points of the feasible region.

Now at these corner points we will find the value of $Z=4x_1+12x_2$

Hence $Z_{(60,0)}=(4×60)+(12×0)=240$

$Z_{(54,24)}=(4×54)+(12×24)=504$

$Z_{(20,40)}=(4×20)+(12×40)=560$

$Z_{(0,20)}=(4×0)+(12×20)=240$

Here we see that at $C(20,40)$ , $Z=4x_1+12x_2$ gives the maximum value.

Therefore $x_1=20$ and $x_2=40$ is the optimal solution of the above model.