Question

If the cost function and revenue function of $x$ items are respectively given as $C(x) = 100 + 0.015x^2$, $R(x) = 3x$, then the value of $x$ for maximum profit is

• A. 50
• B. 100
• C. 150
• D. 200

Hint:

To find the maximum profit, set the derivative of the profit function to zero and solve for $x$.

Answer 1

The Correct Option is B

Profit is defined as $P(x) = R(x) - C(x)$.
Given: $R(x) = 3x$, $C(x) = 100 + 0.015x^2$
So, $P(x) = 3x - (100 + 0.015x^2) = -0.015x^2 + 3x - 100$
To maximize profit, take the derivative and set it to 0:
$\frac{dP}{dx} = -0.03x + 3$
Set $\frac{dP}{dx} = 0 \Rightarrow -0.03x + 3 = 0 \Rightarrow x = \frac{3}{0.03} = 100$
Hence, the profit is maximized when $x = 100$.