A cylindrical can must hold exactly $1500{\text{cm}}^3$ of oil. The material for the top and bottom costs Rs $0.04$ 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) of the can that minimise the total cost. Give your answers correct to two decimal places.

Step-by-step solution

1. Volume constraint: $\pi r^{2}h=1500$ $\Rightarrow$ $h=\frac{1500}{\pi r^2}$.
2. Cost of top and bottom: $2\times 0.04\times \pi r^2=0.08\pi r^2$.
3. Cost of curved side: $0.02\times 2\pi rh=0.04\pi rh$.
4. Total cost $C(r)=0.08\pi r^2+0.04\pi rh$.
5. Substitute $h$ from the volume relation: $C(r)=0.08\pi r^2+0.04\pi r\left(\frac{1500}{\pi r^2}\right)=0.08\pi r^2+\frac{60}{r}$.
6. Differentiate $C(r)$ with respect to $r$: $C^{\prime }(r)=0.16\pi r-\frac{60}{r^2}$.
7. Set derivative to zero for minimum: $0.16\pi r=\frac{60}{r^2}$.
8. Multiply both sides by $r^2$: $0.16\pi r^3=60$.
9. Solve for $r$: $r^3=\frac{60}{0.16\pi }=\frac{60}{0.502654}=119.43$, so $r=\sqrt[3]{119.43}\approx 5.48\text{cm}$.
10. Find $h$ using the volume relation: $h=\frac{1500}{\pi (5.48{)}^2}\approx \frac{1500}{\pi \times 30.02}\approx \frac{1500}{94.34}\approx 15.85\text{cm}$.
11. Check second derivative: $C^{\prime \prime }(r)=0.16\pi +\frac{120}{r^3}>0$, confirming a minimum.

Concept tested: Optimization with a volume constraint

Published 11 September 2026 · How this solution was prepared and checked