On top of a circular cylinder (radius r, height h) we attach a semi-sphere (radius r, center on the cylinder axis). For which values r and h is the surface O of the complete body minimal, if the volume V is given?
I can't imagine such an image even, anyone could help please?
You have a cylinder of base radius $r$ , and height $h$. and a the top you attach a hemi-sphere of radius $r$, so total volume is
$V = \text{constant} = \pi \left( r^2 h + \dfrac{2}{3} r^3 \right) $
The above equation sets a relation between $r$ and $h$.
The total surface area (which we want to minimize) is given by
$ A(r, h) = 2 \pi r h + \pi r^2 + 2 \pi r^2 = \pi ( 2 r h + 3 r^2 ) $
From the equation of the volume we can solve for $h$ in terms of $r$.
$ h = \dfrac{ \left(\dfrac{V}{\pi} \right) - \frac{2}{3} r^3 }{ r^2 } = \dfrac{V}{\pi} \left(\dfrac{1}{r^2} \right) - \dfrac{2}{3} r$
Plug this into the equation of the surface area:
$A(r) = \pi \left( \dfrac{2V}{\pi} \left( \dfrac{1}{r} \right) + \dfrac{5}{3} r^2 \right) $
Now, we just need to differentiate $A(r)$ with respect to $r$
$A'(r) = \pi \left( \dfrac{2V}{\pi} \left( -\dfrac{1}{r^2} \right) + \dfrac{10}{3} r \right ) $
Setting $A'(r) = 0$, and solving for $r$, we get
$ r^3 = \dfrac{3 V}{5 \pi} $
so that the area minimizing $r$ is given by
$ r = \sqrt[3]{\dfrac{3V}{5 \pi}} $
And knowing this, we can determine the corresponding value of the height $h$.
Plug $r$ in the formula for $h$
$h = \dfrac{5}{3} r - \dfrac{2}{3} r = r $