GATE CS 2012 Q56 — General Aptitude

To find the number of units $q$ that maximizes the profit, we first define the profit function $\Pi(q)$ as the difference between total revenue and total cost:

$$\Pi(q) = \text{Revenue} - \text{Cost}$$

Given:

• Market price per unit = $\text{₹ } 50$
• Total Revenue, $R(q) = 50 \times q = 50q$
• Total Cost, $C(q) = 5q^2$

Thus, the profit function is:

$$\Pi(q) = 50q - 5q^2$$

To find the value of $q$ that maximizes $\Pi(q)$, we differentiate $\Pi(q)$ with respect to $q$ and set it to zero:

$$\frac{d\Pi(q)}{dq} = 0$$

$$\frac{d}{dq}(50q - 5q^2) = 50 - 10q = 0$$

$$10q = 50 \implies q = 5$$

To confirm that this point represents a maximum, we check the second derivative:

$$\frac{d^2\Pi(q)}{dq^2} = -10 < 0$$

Since the second derivative is negative, the profit is maximized at $q = 5$ units.