# Maximising the Volume of a Cylinder

A cylinder is to be made out of $12\text{m}^2$ of sheet metal. What must the area of the base be to maximise the volume of the cylinder?

My reasoning is the following:

The cylinder has a base area of $\pi r^2$ and a height of $\displaystyle\frac{6}{\pi r}$. This is because the sheet metal is $12\text{m}^2$ only with the following dimensions:

$~~~~~~~~~~~~~~~~~~~~~~~~~$

Once this net is rolled up, the cylinder is formed with the volume

$$V=\pi r^2\left(\frac{6}{\pi r}\right)$$ $$=6r~~~~~~~~~$$ $$\Rightarrow \frac{\text{d}V}{\text{d}r}=6~~~~~~~~~~~~~~~~~~~~~~~$$

But this means the volume never has a maximum, and is proportional to the value of $r$. Why is this happening?

• See the foil would create an hollow cylinder so you need to use formula of curved surface area also think dimensionally you are doing $m^2=m^3$ – Archis Welankar Mar 2 '16 at 12:57

Surface area of cylinder = $2\pi r^2 + 2\pi r h = 12$ which gives $h = \frac{12}{2\pi r} - r$, now volume is $\pi r^2 h = \pi r^2 \Bigg( \frac{12}{2\pi r} - r \Bigg)$ Now proceed...