Question

The maximum value of Z = 4x + 3y subjected to the constraints 3x + 2y ≥ 160, 5x + 2y ≥ 200, x + 2y ≥ 80; x, y ≥ 0 is

A. 320

B. 300

C. 230

D. None of these

Answer 1

Correct answer is D.

Given object function is

Z = 4x + 3y

Constraints are

3x + 2y ≥ 160

5x + 2y ≥ 200

x + 2y ≥ 80

x ≥ 0

y ≥ 0.

Consider, the inequalities as equalities for some time,

3x + 2y = 160 ; 5x + 2y = 200 and x + 2y = 80

If we convert these into intercept line format equations, we get,

[Dividing the whole equation with the right hand side number of the equation]

From this form of line, we can say that the line 3x + 2y = 160 meets the x-axis at \(\left(\frac{160}{3},0 \right)\) and y-axis at (0, 80).

This shows the inequality 3x + 2y ≥ 160 holds good in the below blue colored region.

Similarly, from the intercept line format, we can say that the line 5x + 2y = 200 meets the x-axis at (40, 0) and y-axis at (0, 100).

This shows the inequality 5x + 2y ≥ 200 holds above in the pink colored region.

Similarly from the intercept line format, we can say that the line x + 2y = 80 meets the x-axis at (80,0) and y-axis at (0, 40).

This shows the inequality x + 2y ≥ 80 holds above in the green colored region.

Now considering the inequalities, x ≥ 0 and y ≥ 0, this clearly shows the region where both x and y are positive. This represents the 1^st quadrant of the graph.

So, the solutions of the LPP are in the first quadrant where the inequalities meet.

Now by plotting all the graphs 3x + 2y ≥ 160 , 5x + 2y ≥ 200 and x + 2y ≥ 80 we get the below graph.

We can clearly see that, there is no area in the 1^st quadrant where all the three inequalities met.

This clearly says that there is no solution for the LPP with the given constraints.