Question

Maximize Z = 5x + 3y
Subject to
$$\begin{gathered} 3x+5y\le 15 \\ 5x+2y\le 10 \\ x,y\ge 0 \end{gathered}$$

Answer 1

First, we will convert the given inequations into equations, we obtain the following equations:

3x + 5y = 15, 5x + 2y = 10, x = 0 and y = 0

Region represented by 3x + 5y ≤ 15 :
The line 3x + 5y = 15 meets the coordinate axes at A(5,0) and B(0,3) respectively. By joining these points we obtain the line 3x + 5y = 15.
Clearly (0,0) satisfies the inequation 3x + 5y ≤ 15. So,the region containing the origin represents the solution set of the inequation 3x + 5y ≤ 15.

Region represented by 5x + 2y ≤ 10 :
The line 5x + 2y = 10 meets the coordinate axes at C(2,0) and D(0,5) respectively. By joining these points we obtain the line 5x + 2y = 10.
Clearly (0,0) satisfies the inequation 5x + 2y ≤ 10. So,the region containing the origin represents the solution set of the inequation 5x + 2y ≤ 10.

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 + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, and y ≥ 0, are as follows.

The corner points of the feasible region are O(0, 0), C(2, 0),$E\left(\frac{20}{19},\frac{45}{19}\right)$ and B(0, 3).

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

Corner point	Z = 5x + 3y
O(0, 0)	5 × 0 + 3 × 0 = 0
C(2, 0)	5 × 2 + 3 × 0 = 10
$E\left(\frac{20}{19},\frac{45}{19}\right)$	5 × $\frac{20}{19}$ + 3 × $\frac{45}{19}$ = $\frac{235}{19}$
B(0, 3)	5 × 0 + 3 × 3 = 9

Therefore, the maximum value of Z is $\frac{235}{19}\mathrm{at}\mathrm{the}\mathrm{point}\left(\frac{20}{19},\frac{45}{19}\right)$.Hence, x= $\frac{20}{19}$ and y =$\frac{45}{19}$ is the optimal solution of the given LPP.
Thus, the optimal value of Z is $\frac{235}{19}$.