Let the total revenue be R(q) and the total cost be C(q).
Selling price per unit = ₹ 50
Total Revenue: $R(q) = 50q$
Total Cost: $C(q) = 5q^2$
The profit function $P(q)$ is:
P(q)=R(q)−C(q)=50q−5q2 To find the value of q that maximizes profit, find the critical points:
dqdP=50−10q=0⟹q=5 Check the second derivative for maximization:
dq2d2P=−10<0 Since the second derivative is negative, profit is maximized at q=5.
Hence, the correct option is (A).