Csdrhrt

I am a bit confused by this problem I have encountered:

A right circular cylindrical container with a closed top is to be constructed
with a fixed surface area. Find the ratio of the height to the
radius which will maximize the volume.

I know the volume to be $ pi{r}^2h$, but I don't see what equation I should be solving for. How can I solve for the ratio?

Thanks

Let $r$ be the radius & $h$ be the height of the cylinder having its total surface area $A$ (constant) since cylindrical container is closed at the top (circular) then its surface area (constantfixed) is given as $$=text{(area of lateral surface)}+2text{(area of circular top/bottom)}$$$$A=2pi rh+2pi r^2$$
$$h=frac{A-2pi r^2}{2pi r}=frac{A}{2pi r}-rtag 1$$

Now, the volume of the cylinder $$V=pi r^2h=pi r^2left(frac{A}{2pi r}-rright)=frac{A}{2}r-pi r^3$$
differentiating $V$ w.r.t. $r$, we get $$frac{dV}{dr}=frac{A}{2}-3pi r^2$$
$$frac{d^2V}{dr^2}=-6pi r<0 (forall r>0)$$
Hence, the volume is maximum, now, setting $frac{dV}{dr}=0$ for maxima $$frac{A}{2}-3pi r^2=0implies color{red}{r}=color{red}{sqrt{frac{A}{6pi}}}$$
Setting value of $r$ in (1), we get
$$color{red}{h}=frac{A}{2pisqrt{frac{A}{6pi}}}-sqrt{frac{A}{6pi}}=left(sqrt{frac{3}{2}}-frac{1}{sqrt 6}right)sqrt{frac{A}{pi}}=color{red}{sqrt{frac{2A}{3pi}}}$$
Hence, the ratio of height $(h)$ to the radius $(r)$ is given as
$$color{}{frac{h}{r}}=frac{sqrt{frac{2A}{3pi}}}{sqrt{frac{A}{6pi}}}=sqrt{frac{12pi A}{3pi A}}=2$$ $$bbox[5pt, border:2.5pt solid #FF0000]{color{blue}{frac{h}{r}=2}}$$

Let the ratio of height to radius be $rho$, then $h=rho r$.

The volume of the cylinder is $V=pi r^2 h=pi rho r^3$.

The surface area of the cylinder with closed ends is

$A=2pi r h + 2pi r^2=2pi rho r^2+2pi r^2=2pi r^2(1+rho)$

hence $r=sqrt{frac{A}{2pi (1+rho)}}$

So the problem is now to find $rho$ which maximises the volume:
$$
V(rho)=pi rho left( frac{A}{2pi (1+rho)}right) ^{3/2}
$$

(which reaches its maximum at $rho=2$)

You can also use Lagrange multipliers. We want the volume to be maximized,
$$ V(r,h)= pi r^{2}h$$
We also have our restraint equation which is
$$S(r,h)= 2pi rh + 2pi r^2 $$
Since we want the surface area to be constant, we have $S(r,h)= k$, k is a number
Now, we can set up equations using Lagrange equations.
$$V_{r}=2pi hr = lambda( 2pi h + 4pi r )......1$$
$$ V_{h}=pi r^{2}=lambda( 2pi r ) .........2$$
$$ k= 2pi rh + 2pi r^2 ..........3$$
We have three equations with 3 unknowns.
Multiply equation 1 by $r$ and equation 2 by $h.$
You get ,
$$ 2pi r^{2}h=lambda 2pi rh + lambda 4pi r^2......4$$
$$ pi r^{2}h=lambda( 2pi rh ) ........5$$
From this multiply EQ.5 by $2$ and then subtract the equation by EQ. 4to get
$$ 2h= 4r$$
$$ frac{h}{r}=2$$

Let $k=h/r$, where $h$ is the height and $r$ is the radius.

The surface of the cylinder is
$$S=2pi r^2+2pi rh=2pi r^2+2pi r^2k=2pi r^2(1+k)$$
Then
$$r=sqrt{frac S{2pi(1+k)}}$$

Now, the volume is
$$V(k)=pi r^2h=pi r^3k=Cfrac k{(k+1)^{3/2}}$$
where $C$ is a constant. Namely, $C=sqrt{frac{S^3pi}8}$.

Now let
$$f(k)=lnfrac{k}{(k+1)^{3/2}}=ln k-frac32ln(k+1)$$
Since $ln$ is increasing and $C$ is positive, clearly $V$ meets its maximum where $f$ does, so let's differentiate $f$:
$$f'(k)=frac1k-frac3{2k+2}=frac{2-k}{k(2k+2)}$$
which vanishes at $k=2$.

The surface area $S=2pi r^2+2pi rh$ is constant,

so $displaystylefrac{dS}{dr}=4pi r+2pi rfrac{dh}{dr}+2pi h=0implies 2r + rfrac{dh}{dr}+h=0$.

When the volume $V=pi r^2h$ is a maximum,

$;;;displaystylefrac{dV}{dr}=pi r^2frac{dh}{dr}+2pi rh=0implies rfrac{dh}{dr}=-2h$.

Therefore the volume is a maximum when $2r-2h+h=0,;$ so $h=2r$ and $displaystylefrac{h}{r}=2$.