A large industrial water tank is made of galvanised steel. It is in the shape of a closed cylinder and has volume of 216000 litres.
The tank has radius $r$ metres and height $h$ metres.
a) Show that the surface area, $Sm^2$, of the tank is given by: $$S=2\pi r^2 + \frac{432}{r}$$
b) i) Find the value of $r$ which minimises the surface area of the cylinder.
ii) Determine the minimum surface area.
I was able to complete part a. On part bi I differentiated(to find the maximum and minimum values of the function)but I do not know how to solve the differentiated equation to get values for $r$. $$S=2\pi r^2 + \frac{432}{r}$$
$$\frac{dS}{dr} = 4\pi r -\frac{432}{r^2}$$
Any help on how to solve this equation would be greatly appreciated. Thanks in advance.
You must equate it to zero, and find the value of $r$.
$$r = {\left(\dfrac{108}{\pi}\right)}^{\frac{1}{3}}$$
If you find $S''$ you'll see it is greater than zero, hence minima.