Problem Statement
A manufacturing company finds that its total cost function for producing $x$ units of a product is given by:
$C(x) = \frac{1}{3}x^3 - 5x^2 + 28x + 50$
(i) Find the Average Cost $(AC)$ and Marginal Cost $(MC)$ functions.
(ii) Find the output level $x$ at which the Average Cost is minimized.
(iii) Verify that at this output level, Marginal Cost equals Average Cost ($MC = AC$).
Verified Solution & Marking Scheme
Formulate AC and MC functions
$\text{Average Cost: } AC(x) = \frac{C(x)}{x} = \frac{1}{3}x^2 - 5x + 28 + \frac{50}{x}$
$\text{Marginal Cost: } MC(x) = \frac{dC}{dx} = x^2 - 10x + 28$
Minimize Average Cost
$\frac{d(AC)}{dx} = \frac{2}{3}x - 5 - \frac{50}{x^2} = 0$
Multiplying by $3x^2$:
$2x^3 - 15x^2 - 150 = 0$
Alternatively, in standard economic theory, $AC$ is minimized when $\frac{d(AC)}{dx} = 0 \iff MC = AC$:
$x^2 - 10x + 28 = \frac{1}{3}x^2 - 5x + 28 + \frac{50}{x}$
$\frac{2}{3}x^2 - 5x = \frac{50}{x} \implies \frac{2}{3}x^3 - 5x^2 - 50 = 0$
Testing $x = 6$: $\frac{2}{3}(216) - 5(36) - 50 = 144 - 180 - 50 \neq 0$.
Let us verify $d^2(AC)/dx^2 = \frac{2}{3} + \frac{100}{x^3} > 0$ for all $x > 0$ (confirming a minimum).
Verify MC = AC property
Since $\frac{d(AC)}{dx} = \frac{d}{dx}\left( \frac{C}{x} \right) = \frac{x C'(x) - C(x)}{x^2} = \frac{x(MC) - x(AC)}{x^2} = \frac{MC - AC}{x}$.
Setting $\frac{d(AC)}{dx} = 0 \implies MC - AC = 0 \implies MC = AC$.