Question

Maximize Z = 3x + 5y
Subject to
$$\begin{gathered} x+2y\le 20 \\ x+y\le 15 \\ y\le 5 \\ x,y\ge 0 \end{gathered}$$

Answer 1

We need to maximize Z = 3x + 5y

First, we will convert the given inequations into equations, we obtain the following equations:
x + 2y = 20, x + y = 15, y = 5 , x = 0 and y = 0.

The line x + 2y = 20 meets the coordinate axis at A(20, 0) and B(0,10). Join these points to obtain the line x + 2y = 20.
Clearly, (0, 0) satisfies the inequation x + 2y ≤ 20. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

The line x + y = 15 meets the coordinate axis at C(15, 0) and D(0,15). Join these points to obtain the line x + y = 15.
Clearly, (0, 0) satisfies the inequation x + y ≤ 15. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

y = 5 is the line passing through (0, 5) and parallel to the X axis.The region below the line y = 5 will satisfy the given inequation.

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.
These lines are drawn using a suitable scale.

The corner points of the feasible region are O(0, 0),$C\left(15,0\right)$, $E\left(10,5\right)$ and $F\left(0,5\right)$

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

Corner point	Z = 3x + 5y
O(0, 0)	3 × 0 + 5 × 0 = 0
$C\left(15,0\right)$	3 × 15 + 5 × 0 = 45
$E\left(10,5\right)$	3 × 10 + 5 × 5 = 55
$F\left(0,5\right)$	3 × 0 + 5 × 5 = 25

We see that the maximum value of the objective function Z is 55 which is at $E\left(10,5\right)$.
Thus, the optimal value of Z is 55.