Question

Refer to question 11. How many of circuits of type A and of type B, should be produced by the manufacturer, so as to maximise his profit? Derermine the maximum profit.

Answer 1

Referring to solution 11, we have
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 constrains is OABCD and is bounded and the coordinates of corner points are (0,0)(10,0), $(\frac{28}{3},\frac{4}{3})$, (6,3) and (0,5), respectivley.
[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]

$\begin{array}{cc}\text{Corner points}& \text{Corresponding value of Z =50x+60y}\\ (0,0)& 0\\ (10,0)& 500\\ (\frac{28}{3},\frac{4}{3})& \frac{1400}{3}+\frac{240}{3}=\frac{1640}{3}=546.66\leftarrow Maximum\\ (6,3)& 480\\ (0,5)& 300\end{array}$

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.