Question

Maximize Z = 50x + 30y
Subject to
$$\begin{gathered} 2x+y\le 18 \\ 3x+2y\le 34 \\ x,y\ge 0 \end{gathered}$$

Answer 1

First, we will convert the given inequations into equations, we obtain the following equations:
2x + y = 18, 3x + 2y = 34

Region represented by 2x + y ≥ 18:
The line 2x + y = 18 meets the coordinate axes at A(9, 0) and B(0, 18) respectively. By joining these points we obtain the line 2x + y = 18.
Clearly (0,0) does not satisfies the inequation 2x + y ≥ 18. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 2x + y ≥ 18.

Region represented by 3x + 2y ≤ 34:
The line 3x + 2y = 34 meets the coordinate axes at $C\left(\frac{34}{3},0\right)$ and D(0, 17) respectively. By joining these points we obtain the line 3x + 2y = 34.
Clearly (0,0) satisfies the inequation 3x + 2y ≤ 34. So,the region containing the origin represents the solution set of the inequation 3x + 2y ≤ 34.

The corner points of the feasible region are A(9, 0), $C\left(\frac{34}{3},0\right)$ and E(2, 14).

The values of Z at these corner points are as follows.

Corner point	Z = 50x + 30y
A(9, 0)	50 × 9 + 3 × 0 = 450
$C\left(\frac{34}{3},0\right)$	50 ×$\frac{34}{3}$ + 30 × 0 = $\frac{1700}{3}$
E(2, 14)	50 × 2 + 30 × 14 = 520

Therefore, the maximum value of Z is $\frac{1700}{3}\mathrm{at}\mathrm{the}\mathrm{point}\left(\frac{34}{3},0\right)$.Hence, x = $\frac{34}{3}$ and y =0 is the optimal solution of the given LPP.
Thus, the optimal value of Z is $\frac{1700}{3}$.