Question

Refer to Exercise 13. Solve the linear programming problem and determine the maximum profit to the manufacturer.

Answer 1

$\begin{array}{cc}\text{Corner Points}& \text{Corresponding value of}X=400x+200y\\ (0,15)& 3000\\ (5,5)& 3000\\ (\frac{30}{7},\frac{30}{7})& 400\times \frac{30}{7}+200\times \times \frac{30}{7}=\frac{18000}{7}\\ & =2571.43\left(minimum\right)\end{array}$

Referring to solution 12, we have minimise Z = 400x + 200y, subject to $5x+2y\ge 30$.

$2x+y\ge 15,x\le y,x\le 0,y\le 0$,

On solving x - y = 0 and 5x + 2y = 30, we get

$y=\frac{30}{7},x=\frac{30}{7}$

On solving x - y = 0 and 2x + y = 15, we get x = 5, y = 5

So, from the shaded feasible region it is clear that coordinates of corner points are (0, 15), (5, 5) and $(\frac{30}{7},\frac{30}{7})$

Hence, the minimum cost is Rs 2571.43.