Question

Maximize Z = x + y
Subject to
$$\begin{gathered} -2x+y\le 1 \\ x\le 2 \\ x+y\le 3 \\ x,y\ge 0 \end{gathered}$$

Answer 1

We need to maximize Z = x + y

First, we will convert the given inequations into equations, we obtain the following equations:
−2x + y = 1, x = 2, x + y = 3, x = 0 and y = 0.

The line −2x + y = 1 meets the coordinate axis at $A\left(\frac{-1}{2},0\right)$ and B(0, 1). Join these points to obtain the line −2x + y = 1 .
Clearly, (0, 0) satisfies the inequation −2x + y ≤ 1. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

x = 2 is the line passing through (2, 0) and parallel to the Y axis.The region below the line x = 2 will satisfy the given inequation.

The line x + y = 3 meets the coordinate axis at C(3, 0) and D(0, 3). Join these points to obtain the line x + y = 3.
Clearly, (0, 0) satisfies the inequation x + y ≤ 3. So, the region in xy-plane that contains the origin represents the solution set of the given equation.

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),$G\left(2,0\right)$, $E\left(2,1\right)$ and $F\left(\frac{2}{3},\frac{7}{3}\right)$

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

Corner point	Z = x + y
O(0, 0)	0 + 0 = 0
$C\left(2,0\right)$	2 + 0 = 2
$E\left(2,1\right)$	2 +1 = 3
$F\left(\frac{2}{3},\frac{7}{3}\right)$	$\frac{2}{3}+\frac{7}{3}=\frac{9}{3}=3$

We see that the maximum value of the objective function Z is 3 which is at $E\left(2,1\right)$ and $F\left(\frac{2}{3},\frac{7}{3}\right)$.
Thus, the optimal value of Z is 3.