Height/Radius ratio for maximum volume cylinder of given surface area 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
 A: 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 (constant\fixed) is given as $$=\text{(area of lateral surface)}+2\text{(area of circular top/bottom)}$$$$A=2\pi rh+2\pi r^2$$ 
$$h=\frac{A-2\pi r^2}{2\pi r}=\frac{A}{2\pi r}-r\tag 1$$
Now, the volume of the cylinder $$V=\pi r^2h=\pi r^2\left(\frac{A}{2\pi r}-r\right)=\frac{A}{2}r-\pi r^3$$
differentiating $V$ w.r.t. $r$, we get $$\frac{dV}{dr}=\frac{A}{2}-3\pi r^2$$
$$\frac{d^2V}{dr^2}=-6\pi r<0\ \ (\forall\ \ r>0)$$
Hence, the volume is maximum, now, setting $\frac{dV}{dr}=0$ for maxima $$\frac{A}{2}-3\pi r^2=0\implies \color{red}{r}=\color{red}{\sqrt{\frac{A}{6\pi}}}$$ 
Setting value of $r$ in (1), we get 
$$\color{red}{h}=\frac{A}{2\pi\sqrt{\frac{A}{6\pi}}}-\sqrt{\frac{A}{6\pi}}=\left(\sqrt{\frac{3}{2}}-\frac{1}{\sqrt 6}\right)\sqrt{\frac{A}{\pi}}=\color{red}{\sqrt{\frac{2A}{3\pi}}}$$
Hence, the ratio of height $(h)$ to the radius $(r)$ is given as 
$$\color{}{\frac{h}{r}}=\frac{\sqrt{\frac{2A}{3\pi}}}{\sqrt{\frac{A}{6\pi}}}=\sqrt{\frac{12\pi A}{3\pi A}}=2$$ $$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{\frac{h}{r}=2}}$$
A: 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 
$$S(r,h)= 2\pi rh + 2\pi r^2 $$
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.
$$V_{r}=2\pi hr = \lambda( 2\pi h + 4\pi r )......1$$
$$ V_{h}=\pi r^{2}=\lambda( 2\pi r ) .........2$$
$$ k= 2\pi rh + 2\pi r^2 ..........3$$
We have three equations with 3 unknowns. 
Multiply equation 1 by $r$ and equation 2 by $h.$ 
You get , 
$$ 2\pi r^{2}h=\lambda 2\pi rh + \lambda 4\pi r^2......4$$
$$ \pi r^{2}h=\lambda( 2\pi rh ) ........5$$
From this multiply  EQ.5 by $2$ and then subtract the equation by EQ. 4to get
$$ 2h= 4r$$
$$ \frac{h}{r}=2$$
A: The surface area $S=2\pi r^2+2\pi rh$ is constant, 
so $\displaystyle\frac{dS}{dr}=4\pi r+2\pi r\frac{dh}{dr}+2\pi h=0\implies 2r + r\frac{dh}{dr}+h=0$.
When the volume $V=\pi r^2h$ is a maximum, 
$\;\;\;\displaystyle\frac{dV}{dr}=\pi r^2\frac{dh}{dr}+2\pi rh=0\implies r\frac{dh}{dr}=-2h$.
Therefore the volume is a maximum when $2r-2h+h=0,\;$ so $h=2r$ and $\displaystyle\frac{h}{r}=2$.
A: 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
$A=2\pi r h + 2\pi r^2=2\pi \rho r^2+2\pi r^2=2\pi r^2(1+\rho)$
hence $r=\sqrt{\frac{A}{2\pi (1+\rho)}}$
So the problem is now to find $\rho$ which maximises the volume:
$$
V(\rho)=\pi \rho \left( \frac{A}{2\pi (1+\rho)}\right) ^{3/2}
$$
(which reaches its maximum at $\rho=2$)
A: Let $k=h/r$, where $h$ is the height and $r$ is the radius.
The surface of the cylinder is
$$S=2\pi r^2+2\pi rh=2\pi r^2+2\pi r^2k=2\pi r^2(1+k)$$
Then
$$r=\sqrt{\frac S{2\pi(1+k)}}$$
Now, the volume is
$$V(k)=\pi r^2h=\pi r^3k=C\frac k{(k+1)^{3/2}}$$
where $C$ is a constant. Namely, $C=\sqrt{\frac{S^3\pi}8}$.
Now let 
$$f(k)=\ln\frac{k}{(k+1)^{3/2}}=\ln k-\frac32\ln(k+1)$$
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 $k=2$.
