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

(d) none of these

We need to maximize the function Z = 4x + 3y

Converting the given inequations into equations, we obtain

$3x+2y=160,5x+2y=200,x+2y=80,x=0\mathrm{and}y=0$

Region represented by 3x + 2y ≥ 160:
The line 3x + 2y = 160 meets the coordinate axes at $A\left(\frac{160}{3},0\right)$ and B(0, 80) respectively. By joining these points we obtain the line
3x + 2y = 160.Clearly (0,0) does not satisfies the inequation 3x + 2y ≥ 160. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 3x + 2y ≥ 160.

Region represented by 5x +2y ≥ 200:
The line 5x +2y = 200 meets the coordinate axes at C(40,0) and D(0, 100) respectively. By joining these points we obtain the line
5x +2y = 200.Clearly (0,0) does not satisfies the inequation 5x +2y ≥ 200. So,the region which does not contain the origin represents the solution set of the inequation 5x +2y ≥ 200.

Region represented by x +2y ≥ 80:
The line x +2y = 80 meets the coordinate axes at E(80,0) and F(0, 40) respectively. By joining these points we obtain the line
x +2y = 80.Clearly (0,0) does not satisfies the inequation x +2y ≥ 80. So,the region which does not contain the origin represents the solution set of the inequation x +2y ≥ 80.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 3x + 2y ≥ 160, 5x +2y ≥ 200, x +2y ≥ 80, x ≥ 0, and y ≥ 0 are as follows.

Here, we see that the feasible region is unbounded. Therefore,maximum value is infinity.