Linear Programming — Class 12 Maths Solution

Referring to Solution 11 , We have the following conditions as per the question,
Maximise $Z = 50x + 60y$,

subject to
$2x + y \le 20,$ $x + 2y \le 12,$ $x + 3y \le 15,$ $x \ge 0,$ $y \ge 0$

From the shaded region it is clear that the feasible region determined by the system of constraints is OABCD and is bounded and the coordinates of corner points are (0,0),

$(10,0),\left( {\frac{{28}}{3},\frac{4}{3}} \right)$,(6,3) and (0,5), respectively.

[since, $x + 2y = 12$ and $2x + y = 20 \Rightarrow x = \frac{{28}}{3},y = \frac{4}{3}$ and $x + 3y = 15$ and $x + 2y = 12 \Rightarrow y = 3$ and $x = 6$]

Since, the manufacturer is required to produce two types of
circuits A and B and it is clear that parts of resistor, transistor and capacitor cannot be in fraction, so the required maximum profit is 480 where circuits of type A is 6 and circuits of type B is 3.