A cylindrical can must hold exactly $2000$ cm${}^3$ of oil. The material for the top and bottom costs Rs$,0.05$ 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. Give your answers correct to two decimal places.

Step-by-step solution

1. Write the volume constraint $\pi r^{2}h=2000$ so $h=\frac{2000}{\pi r^2}$.
2. Write the total cost $C=0.05(2\pi r^2)+0.02(2\pi rh)=0.1\pi r^2+0.04\pi rh$.
3. Substitute $h$ into $C$ to obtain $C(r)=0.1\pi r^2+\frac{80}{r}$.
4. Differentiate: $C^{\prime }(r)=0.2\pi r-\frac{80}{r^2}$.
5. Set $C^{\prime }(r)=0$: $0.2\pi r^3=80⟹r^3=\frac{400}{\pi }$.
6. Solve for $r$: $r=(\frac{400}{\pi }{)}^{1/3}\approx 5.01$ cm.
7. Use the volume relation to find $h=\frac{2000}{\pi r^2}\approx 25.38$ cm.
8. Check the second derivative $C^{\prime \prime }(r)=0.2\pi +\frac{160}{r^3}>0$ to confirm a minimum.

Concept tested: Optimization of cylindrical cost

Published 12 September 2026 · How this solution was prepared and checked