Answer to Question #229617 in Calculus for Muzamil

Question

Determine the price p which result in maximum total revenue

Answer 1

$p=500-\frac{1}{50}q,\\ \text{where p=price and q=demand}\\ R(q)=p×q=500q-\frac{1}{50}q^2, text{where R = revenue}\\ \text{And, maximum revenue is given by derivative of Revenue function}\\ R'(q)=500-\frac{2}{50}q\\ 0=500-\frac{2}{50}q\\ q=12500\\ p=250 \text{Therefore, Revenue R is maximum when price p=250.}$