Question 6 Consider the cost function \( C(x)=5 x^{3}-\frac{4}{x^{2}} \). What is the marginal cost at the production level \( x=2 \) ? Ouestion 7

To find the marginal cost at the production level \( x=2 \) for the cost function \( C(x)=5x^{3}-\frac{4}{x^{2}} \), we need to find the derivative of the cost function with respect to \( x \) and then evaluate it at \( x=2 \). The marginal cost is the rate of change of the cost with respect to the production level, which is given by the derivative of the cost function. Let's find the derivative of the cost function \( C(x) \) with respect to \( x \) using the power rule for differentiation: \[ C'(x) = \frac{d}{dx} \left(5x^{3}-\frac{4}{x^{2}}\right) \] Using the power rule for differentiation, we have: \[ C'(x) = 15x^{2} + \frac{8}{x^{3}} \] Now, we need to evaluate the marginal cost at \( x=2 \) by substituting \( x=2 \) into the derivative: \[ C'(2) = 15(2)^{2} + \frac{8}{(2)^{3}} \] \[ C'(2) = 15(4) + \frac{8}{8} \] \[ C'(2) = 60 + 1 \] \[ C'(2) = 61 \] Therefore, the marginal cost at the production level \( x=2 \) is 61.