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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

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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 $$\color{}{\frac{h}{r}}=\frac{\sqrt{\frac{2A}{3\pi}}}{\sqrt{\frac{A}{6\pi}}}=\sqrt{\frac{12\pi A}{3\pi A}}=2$$ $$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{\frac{h}{r}=2}}$$

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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 $$ 2h= 4r$$ $$ \frac{h}{r}=2$$

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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 $\rho=2$)

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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 $k=2$.

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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 $\displaystyle\frac{h}{r}=2$.

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