Question

Maximize Z = 10x + 6y
Subject to
$$\begin{gathered} 3x+y\le 12 \\ 2x+5y\le 34 \\ x,y\ge 0 \end{gathered}$$

Answer 1

First, we will convert the given inequations into equations, we obtain the following equations:
3x + y = 12, 2x + 5y = 34, x = 0 and y = 0

Region represented by 3x + y ≤ 12:
The line 3x + y = 12 meets the coordinate axes at $A\left(4,0\right)$ and $B\left(0,12\right)$ respectively. By joining these points we obtain the line 3x + y = 12.
Clearly (0,0) satisfies the inequation 3x + y ≤ 12. So,the region containing the origin represents the solution set of the inequation 3x + y ≤ 12 .

Region represented by 2x + 5y ≤ 34:
The line 2x + 5y = 34 meets the coordinate axes at $C\left(17,0\right)$ and $D\left(0,\frac{34}{5}\right)$ respectively. By joining these points we obtain the line 2x + 5y ≤ 34.
Clearly (0,0) satisfies the inequation 2x + 5y ≤ 34. So,the region containing the origin represents the solution set of the inequation 2x + 5y ≤ 34.

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 + y ≤ 12, 2x + 5y ≤ 34, x ≥ 0, and y ≥ 0 are as follows.

The corner points of the feasible region are O(0, 0), $A\left(4,0\right)$ ,$E\left(2,6\right)$ and $D\left(0,\frac{34}{5}\right)$.

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

Corner point	Z = 10x + 6y
O(0, 0)	10 × 0 + 6 × 0 = 0
$A\left(4,0\right)$	10× 4 + 6 × 0 = 40
$E\left(2,6\right)$	10 × 2 + 6 × 6 = 56
$D\left(0,\frac{34}{5}\right)$	10 × 0 + 6 ×$\frac{34}{5}$ = $\frac{204}{3}$

We see that the maximum value of the objective function Z is 56 which is at $E\left(2,6\right)$ that means at x = 2 and y = 6.
Thus, the optimal value of Z is 56.