Question

Minimize Z = 30x + 20y
Subject to
$$\begin{gathered} x+y\le 8 \\ x+4y\ge 12 \\ 5x+8y=20 \\ x,y\ge 0 \end{gathered}$$

Answer 1

First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 8, x + 4y = 12, x = 0 and y = 0

5x + 8y = 20 is already an equation.

Region represented by x + y ≤ 8:
The line x + y = 8 meets the coordinate axes at A(8, 0) and B(0, 8) respectively. By joining these points we obtain the line x + y = 8.Clearly (0,0) satisfies the inequation x + y ≤ 8. So,the region in xy plane which contain the origin represents the solution set of the inequation x + y ≤ 8.

Region represented by x + 4y ≥ 12:
The line x + 4y = 12 meets the coordinate axes at C(12, 0) and D(0, 3) respectively. By joining these points we obtain the line x + 4y = 12.Clearly (0,0) satisfies the inequation x + 4y ≥ 12. So,the region in xy plane which does not contain the origin represents the solution set of the inequation x + 4y ≥ 12.

The line 5x + 8y = 20 is the line that passes through E(4, 0) and $F\left(0,\frac{5}{2}\right)$ .

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 ≤ 8, x + 4y ≥ 12, 5x + 8y = 20, x ≥ 0 and y ≥ 0 are as follows.



The corner points of the feasible region are B(0,8), D(0,3), $G\left(\frac{20}{3},\frac{4}{3}\right)$.
The values of Z at these corner points are as follows.

Corner point	Z = 30x + 20y
B(0,8)	160
D(0,3)	60
$G\left(\frac{20}{3},\frac{4}{3}\right)$	266.66

Therefore, the minimum value of Z is 60 at the point D(0,3). Hence, x = 0 and y =3 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 60.