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?
you should proceed like this:
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...