A company’s total cost (in rupees) for producing $x$ thousand units is given by $C(x)=2000+80x+3x^2$. Find the production level $x$ that minimises the average cost per unit and state the minimum average cost (in rupees per unit).

Step-by-step solution

1. Average cost $A(x)$ is total cost divided by the number of units produced: $A(x)=\frac{C(x)}{x}=\frac{2000+80x+3x^2}{x}=\frac{2000}{x}+80+3x$.
2. Differentiate $A(x)$ with respect to $x$: $A^{\prime }(x)=-\frac{2000}{x^2}+3$.
3. Set the derivative to zero for a minimum: $-\frac{2000}{x^2}+3=0\Rightarrow 3=\frac{2000}{x^2}$.
4. Solve for $x^2$: $x^2=\frac{2000}{3}$, so $x=\sqrt{\frac{2000}{3}}\approx 25.819$ (thousand units).
5. Since $x$ represents thousands of units, the production level is $25.819\times 1000\approx 2582$ units.
6. Compute the minimum average cost by substituting $x$ back into $A(x)$:
7. $A_{\min }=\frac{2000}{25.819}+80+3(25.819)\approx 77.47+80+77.46\approx 234.93$ rupees per unit.
8. Round to the nearest rupee: Rs $235$ per unit.

Concept tested: Minimising average cost using derivatives

Published 15 September 2026 · How this solution was prepared and checked