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, denoted by $P(x)$, is calculated as revenue $R(x)$ minus cost $C(x)$, where $P(x) = R(x) - C(x)$.
Given the revenue function $R(x) = 3x$ and the cost function $C(x) = 100 + 0.015x^2$, the profit function is $P(x) = 3x - (100 + 0.015x^2)$, which simplifies to $P(x) = -0.015x^2 + 3x - 100$.
To find the maximum profit, we compute the derivative of $P(x)$ with respect to $x$ and set it to zero:
$\frac{dP}{dx} = -0.03x + 3$
Setting the derivative to zero: $\frac{dP}{dx} = 0 \Rightarrow -0.03x + 3 = 0$. Solving for $x$ gives $x = \frac{3}{0.03} = 100$.
Therefore, the profit is maximized when $x = 100$.