Q.The total cost function of a firm is $C(x) = 2x^2 + 3x + 50$. Find the output at which average cost is minimum, and verify using the condition $MC=AC$.

Form $AC(x)=C(x)/x$, minimise it directly using calculus, then cross-check against the $MC=AC$ theorem.

Step 1 — Form the average cost function. $AC(x) = \dfrac{C(x)}{x} = \dfrac{2x^2+3x+50}{x} = 2x+3+\dfrac{50}{x}$.

Step 2 — Differentiate and set to zero (calculus method). $\dfrac{d(AC)}{dx} = 2 - \dfrac{50}{x^2}$. Setting this to zero: $2 = \dfrac{50}{x^2} \Rightarrow x^2 = 25 \Rightarrow x=5$ (rejecting $x=-5$, since output cannot be negative).

Step 3 — Confirm it is a minimum. $\dfrac{d^2(AC)}{dx^2} = \dfrac{100}{x^3}$, which is positive for $x=5>0$ — confirming a minimum.

Step 4 — Find the minimum average cost. $AC(5) = 2(5)+3+\dfrac{50}{5} = 10+3+10=23$. …

Whenever a cost function is quadratic, $AC(x)=C(x)/x$ always has the shape $ax+b+c/x$, which by the AM-GM-style minimum of $ax$ and $c/x$ is minimised where $ax=c/x$, i.e. $x=\sqrt{c/a}$ — a quick mental-arithmetic sho …

Minimising the total cost function $C(x)$ instead of the average cost function $AC(x)$ is a common confusion — a quadratic total cost function with positive leading coefficient has no genuine minimum over $x>0$ other than at $x=0$ itself, which is no …