A cylindrical can must hold exactly $2500{\text{cm}}^3$ of oil. The material for the top and bottom costs Rs $0.06$ per cm${}^2$ and the material for the curved side costs Rs $0.02$ per cm${}^2$. Find the radius $r$ and height $h$ (in cm) that minimise the total cost, and state the minimum cost (in rupees). Give $r$ and $h$ correct to two decimal places.

Step-by-step solution

1. Let $r$ be the radius and $h$ the height. Volume constraint: $\pi r^{2}h=2500$, so $h=\frac{2500}{\pi r^2}$.
2. Cost of top and bottom: $2(\pi r^2)(0.06)=0.12\pi r^2$.
3. Cost of curved side: $2\pi rh(0.02)=0.04\pi rh$.
4. Total cost $C(r)=0.12\pi r^2+0.04\pi rh$.
5. Substitute $h$ from the volume condition: $C(r)=0.12\pi r^2+0.04\pi r\left(\frac{2500}{\pi r^2}\right)=0.12\pi r^2+\frac{100}{r}$.
6. Differentiate with respect to $r$: $C^{\prime }(r)=0.24\pi r-\frac{100}{r^2}$.
7. Set $C^{\prime }(r)=0$: $0.24\pi r=\frac{100}{r^2}\Rightarrow 0.24\pi r^3=100$.
8. Solve for $r$: $r^3=\frac{100}{0.24\pi }\approx 132.629$, so $r=(132.629{)}^{1/3}\approx 5.09$cm.
9. Find $h$ using the volume relation: $h=\frac{2500}{\pi (5.09{)}^2}\approx 30.71$cm.
10. Compute minimum cost: $C_{\min }=0.12\pi (5.09{)}^2+\frac{100}{5.09}\approx 9.77+19.65=29.42$ rupees.
11. Thus the optimal dimensions are $r\approx 5.09$cm, $h\approx 30.71$cm, and the minimum cost is about Rs $29.42$.

Concept tested: Optimization – cost minimisation with volume constraint

Published 17 September 2026 · How this solution was prepared and checked