Question

Minimize Z = x − 5y + 20
Subject to
$$\begin{gathered} x-y\ge 0 \\ -x+2y\ge 2 \\ x\ge 3 \\ y\le 4 \\ x,y\ge 0 \end{gathered}$$

Answer 1

First, we will convert the given inequations into equations, we obtain the following equations:
x − y = 0, − x + 2y = 2, x = 3, y = 4, x = 0 and y = 0.

Region represented by x − y ≥ 0 or x ≥ y:
The line x − y = 0 or x = y passes through the origin.The region to the right of the line x = y will satisfy the given inequation.
Let's check by taking an example like if we take a point (4, 3) to the right of the line x = y .Here x ≥ y.So, it satisfy the given inequation. Take a point (4, 5) to the left of the line x = y. Here, x ≤ y. That means it does not satisfy the given inequation.

Region represented by − x + 2y ≥ 2:
The line − x + 2y = 2 meets the coordinate axes at A(−2, 0) and B(0, 1) respectively. By joining these points we obtain the line
− x + 2y = 2.Clearly (0,0) does not satisfies the inequation − x + 2y ≥ 2. So,the region in xy plane which does not contain the origin represents the solution set of the inequation − x + 2y ≥ 2 .

The line x = 3 is the line that passes through the point (3, 0) and is parallel to Y axis. x ≥ 3 is the region to the right of the line x = 3.

The line y = 4 is the line that passes through the point (0, 4) and is parallel to X axis. y ≤ 4 is the region below the line y = 4.

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 x − y ≥ 0,− x + 2y ≥ 2, x ≥ 3, y ≤ 4, x ≥ 0 and y ≥ 0 are as follows.



The corner points of the feasible region are $C\left(3,\frac{5}{2}\right)$,$D\left(3,3\right)$, E(4, 4) and F(6, 4).
The values of Z at these corner points are as follows.

Corner point	Z = x − 5y + 20
$C\left(3,\frac{5}{2}\right)$	3 − 5 × $\frac{5}{2}$ + 20 = $\frac{21}{2}$
$D\left(3,3\right)$	3 − 5 × 3 + 20 = 8
E(4, 4)	4 − 5 × 4 + 20 = 4
F(6, 4)	6 − 5 × 4 + 20 = 6

Therefore, the minimum value of Z is 4 at the point E(4, 4). Hence, x = 4 and y = 4 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 4.