Linear Programming — Class 12 Maths Solution

Referring to Solution 12, We have the following conditions as per the question, minimise $Z = 400x + 200y$, subject to $5x + 2y \ge 30$,

$2x + y \le 15,$ $x \le y,$ $x \ge 0,$ $y \ge 0$
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 $\left( {\frac{{30}}{7}} \right.$,

$\left. {\frac{{30}}{7}} \right)$

Hence, the minimum cost is Rs.2571.43.