A cylindrical can is to hold exactly $500{\text{cm}}^3$ of liquid. The material for the top and bottom costs $\text{Rs}0.02$ per ${\text{cm}}^2$, while the material for the curved side costs $\text{Rs}0.01$ per ${\text{cm}}^2$. 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 $r$ be the radius (cm) and $h$ the height (cm) of the cylinder.
2. Volume constraint: $\pi r^{2}h=500\Rightarrow h=\frac{500}{\pi r^2}$.
3. Surface area of top and bottom: $2\pi r^2$; cost for these parts: $0.02\times 2\pi r^2=0.04\pi r^2$ rupees.
4. Lateral surface area: $2\pi rh$; cost for the side: $0.01\times 2\pi rh=0.02\pi rh$ rupees.
5. Total cost function: $C(r)=0.04\pi r^2+0.02\pi rh$.
6. Replace $h$ using the volume relation: $C(r)=0.04\pi r^2+0.02\pi r\left(\frac{500}{\pi r^2}\right)=0.04\pi r^2+\frac{10}{r}$.
7. Differentiate $C(r)$ with respect to $r$: $C^{\prime }(r)=0.08\pi r-\frac{10}{r^2}$.
8. Set derivative to zero for minimum: $0.08\pi r-\frac{10}{r^2}=0\Rightarrow 0.08\pi r^3=10$.
9. Solve for $r$: $r^3=\frac{10}{0.08\pi }=\frac{125}{\pi }$, so $r={\left(\frac{125}{\pi }\right)}^{1/3}$.
10. Numerically, $\frac{125}{\pi }\approx 39.79$, and its cube root $\approx 3.42\text{cm}$.
11. Find $h$ using $h=\frac{500}{\pi r^2}$: $r^2\approx 11.70$, $\pi r^2\approx 36.73$, hence $h\approx \frac{500}{36.73}\approx 13.61\text{cm}$.
12. Thus the cost‑effective dimensions are $r\approx 3.42\text{cm}$ and $h\approx 13.61\text{cm}$.

Concept tested: Optimization – cost minimisation

Published 6 September 2026 · How this solution was prepared and checked