Question

The total cost function for x units of a commodity is given by \(C(x) = \frac{25x^3}{3}-75x^2+48x+34\). The output \(x\) at which the marginal cost is minimum is:

• A. 6
• B. 3
• C. 10
• D. 5

Answer 1

The Correct Option is B

To find the output \(x\) at which the marginal cost is minimum, we first need to determine the marginal cost function. The marginal cost \(MC(x)\) is the derivative of the total cost function \(C(x)\) with respect to \(x\). Given:

\(C(x) = \frac{25x^3}{3} - 75x^2 + 48x + 34\)

Let's differentiate \(C(x)\) with respect to \(x\):

\(MC(x) = \frac{d}{dx} \left(\frac{25x^3}{3} - 75x^2 + 48x + 34\right)\)

Using power rule for derivatives, we have:

\(MC(x) = 25x^2 - 150x + 48\)

To find the minimum marginal cost, we need to find the critical points by setting the derivative of \(MC(x)\) to zero:

\(MC'(x) = \frac{d}{dx}(25x^2 - 150x + 48) = 50x - 150\)

Set \(MC'(x) = 0\):

\(50x - 150 = 0\)

Solve for \(x\):

\(50x = 150\)

\(x = 3\)

To confirm this is a minimum, we check the second derivative \(MC''(x)\):

\(MC''(x) = \frac{d}{dx}(50x - 150) = 50\)

Since \(MC''(x) = 50 > 0\), the function is concave up at \(x = 3\) and therefore, the marginal cost is minimized at \(x = 3\). Thus, the output \(x\) at which the marginal cost is minimum is 3.