A radio manufacturer produces x sets per week at a total cost of $₹\begin{Bmatrix} \frac{x^2}{25}+3x+100\end{Bmatrix}$. He is a monopolist and the demand of his product is $x=75-3p $, where p is the price in rupees per se t. For maximum net revenue the number of sets produced per week is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 30
Demand function, $x=75-3p$
$⇒p=\frac{x-75}{3}$
∴ Revenue, $R(x)=x.p=\frac{x×(75-x)}{3}=\frac{75x-x^2}{3}$
Now,
Profit = $R(x)-C(x)$
$=\frac{75x-x^2}{3}-\left(\frac{x^2}{25}+3x+100\right)$
$=\frac{1650x-28x^2}{75}-100$
for critical point, $P'(c)=0$
$⇒1650-56x=0$
$⇒x=\frac{1650}{56}≃30$