A cylindrical can is to hold exactly $2000{\text{ cm}}^3$ of liquid. The material for the top and bottom costs Rs $0.03$ per cm${}^2$, while the material for the curved side costs Rs $0.015$ per cm${}^2$. Find the radius and height of the can that minimise the total cost. Give your answer correct to two decimal places.

Step-by-step solution

1. Let radius be $r$ cm and height be $h$ cm. Volume constraint: $V=\pi r^{2}h=2000$. Express $h=\frac{2000}{\pi r^2}$.
2. Total cost $C$ is sum of cost of top and bottom: $2cdot0.03cdotpi{r}^2=0.06pi{r}^2$, and cost of lateral surface: $0.015cdot(2pirh)=0.03pirh$. So $C=0.06pi{r}^2+0.03pirh$.
3. Substitute $h$ into $C$: $C=0.06\pi r^2+0.03\pi rcdot\frac{2000}{\pi r^2}=0.06\pi r^2+\frac{60}{r}$.
4. Differentiate $C$ w.r.t. $r$: $C^{\prime }(r)=0.12\pi r-\frac{60}{r^2}$.
5. Set $C^{\prime }(r)=0$ and solve: $0.12\pi r=\frac{60}{r^2}\Rightarrow 0.12\pi r^3=60\Rightarrow r^3=\frac{60}{0.12\pi }=\frac{500}{\pi }$. Thus $r={\left(\frac{500}{\pi }\right)}^{1/3}$.
6. Compute numeric value: $\frac{500}{\pi }\approx 159.155$, cube root $\approx 6.80\text{ cm}$.
7. Find $h$ from $h=\frac{2000}{\pi r^2}$. With $r=6.80$, $r^2\approx 46.24$, so $happrox\frac{2000}{picdot46.24}\approx 10.92\text{ cm}$.
8. Second derivative $C^{\prime \prime }(r)=0.12\pi +\frac{120}{r^3}>0$ for $r>0$, confirming a minimum.

Concept tested: Optimization – cost minimisation of a cylinder

Published 8 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked