Optimization with cylinder I have no idea how to do this problem at all.
A cylindrical can without a top is made to contain V cm^3 of liquid. Find the dimensions that will minimize the cost of the metal to make the can.
Since no specific volume is given the smallest amount of metal for the can would be zero, which would held zero cm^3 of liquid. How is this wrong? It is not possible to make a cylinder out of a negative amount of metal.
 A: In the cylinder without top, the volume $V$ is given by:
$$V=\pi R^2h$$
the surface,
$$S=2\pi Rh+\pi R^2$$
Solving the first eq. respect to $R$, you find:
$$h=\frac{V}{\pi R^2}$$
Putting this into the equation of the surface, you obtaine:
$$S=2\frac{V}{R}+\pi R^2$$
deriving this expression respect to $R$ and putting the result to zero in order to find the minimum, you have:
$$R=\sqrt[3]\frac{V}{\pi}$$
A: As an alternative to @Riccardo.Alestra's fine answer,
and with a discussion of proper treatment of critical points
(to justify that the solution is indeed a global minimum)...
We wish to minimize $\pi(r^2+2rh)$ subject to $\pi r^2h=V$. Without preference for either $r$ or $h$, we could proceed using differentials. The constraint becomes a relation between $dr,dh$:
$$
0 = dV = \pi \cdot d\left(r^2h\right)
$$
or, dispensing with the multiples of $\pi$ and then $r$,
$$
0 = r^2 dh + 2rh \, dr \implies
$$
$$
0 = r \, dh + 2h \, dr \implies
$$
$$
\frac{dr}{dh}=-\frac12\frac{r}{h}
\qquad
\text{or}
\qquad
\frac{dh}{dr}=-2\frac{h}{r}
\,.
$$
Now we turn to the objective function, also 
dispensing with the constant multiples of $\pi$ and then $2$:
$$
\frac{A}{\pi} = r^2+2rh
$$
$$
\eqalign{
0
&= d\left( r^2+2rh \right) \\
&= 2r\,dr + 2\left( r\,dh + h \, dr \right)
\implies\\
0
&= r\,dr + r\,dh + h \, dr \\
&= \left(r+h\right)\,dr + r\,dh
}
$$
At this point, we use one of the two equivalent differential ratios above:
$$
\eqalign{
0
&= r+h + r\,\frac{dh}{dr} \\
&= r+h + r\,\left(-2\frac{h}{r}\right) \\
&= r+h -2 r\,\frac{h}{r} \\
&= r+h -2 h \\
&= r-h \\\\
&\iff\qquad r=h
}
$$
Putting this back into the constraint
(and being forced to prefer one variable, say $r$),
we obtain
$$
\eqalign{
V &= \pi r^3 = \pi h^3 \\
r &= h = \left(\frac{V}{\pi}\right)^{1/3} \\
}
$$
Lastly, we need to ensure that this is a global minimum
and not a local minimum or global or local maximum.
To see this, we either need the second derivative of our objective function
$f$ or else a numberline sketch of the sign of $f\,'$ for $r,h>0$ (satisfying the constraint, which should also be graphed to see the inverse relationship).
Recall that our objective function
$$
f(r)=\pi\left(r^2+2rh\right)
$$
has derivative
$$
f\,'(r)=\pi r\left(r-h\right)
$$
which is negative for $r\in(0,h)$ and positive for $r > h$,
so that $r=h$ is indeed the global minimum.
One can also, of course, compute
$$
\eqalign{
f\,''(r)
&= \pi \, \frac{d}{dr} \left( r^2 - rh \right) \\
&= \pi \, \left( 2r - h - r \, \frac{dh}{dr} \right) \\
&= \pi \, \left( 2r + h \right) \implies \\
\Bigl. f\,''(r) \Bigr|_{r=h} &= 3\pi r > 0
}
$$
which shows that $r=h$ is at least a local minimum,
but we must observe that $f\,''>0$ for all $r,h>0$
(i.e. that $f$ is strictly concave)
to conclude that it is in fact a global minimum.
A: The volume of a cylindrical can is given by $\pi r^2h$, where $r$ is the radius of the base and $h$ is the height. The area of the surface is given by: $2\pi rh$(-area of the side)+$\pi r^2$(-area of the bottom), there is no top.
From the given $V$, you can express $h=\frac{V}{\pi r^2}$. Substitute to the second equation to get $S(r)=\frac{2V}{r}+\pi r^2$. This is a function of one variable, $r$. Find it's minimum by differentiation. $S'(r)=-\frac{2V}{r^2}+2\pi r=\frac{2\pi r^3-2V}{r^2}$. Now, for $S'(r)=0$, we need $2\pi r^3-2V=0$, so $r=\sqrt[3]\frac{V}{\pi}$.
