# 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.

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$V$ is given, although not numerically. Write equations for the volume and area of the cylinder in terms of the radius and height. Solve for those variables, using $V$ as a constant. – lhf Apr 3 '12 at 10:55
But I don't understand how 0 isn't the answer. – user138246 Apr 3 '12 at 11:14
Because a cylinder with zero area has zero volume. So unless $V=0$, the answer cannot be $0$. – lhf Apr 3 '12 at 11:21
The problem doesn't specify that the volume is > 0. – user138246 Apr 3 '12 at 11:22
@Jordan: EXACTLY. The problem doesn't specify what $V$ is! You are trying to specify that it is zero. Why not let go of that faulty assumption and proceed with the calculus? It's not a trick question... – The Chaz 2.0 Apr 3 '12 at 13:12

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:

$$R=\sqrt[3]\frac{V}{\pi}$$

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Note that $V = \pi R^2h$ was solved for $h$, not for $R$. – Johannes Kloos Apr 3 '12 at 11:34
Jordan, in light of the chat a few hours ago, let me comment on your approach to this problem. The problem explicitly mentions the VOLUME OF A CYLINDER. As much as you want this to be a trick question, you will not get the solution unless you write THE FORMULA for the volume of a cylinder!!! – The Chaz 2.0 Apr 3 '12 at 13:06
Then when it mentions "the metal [used] to make the can", the surface area is implied. Observe that there is no top to this cylinder, so adjust the surface area formula appropriately. Then you need to take the derivative of something (because you're in a calculus class!), and it's best to use substitutions as Riccardo has done. – The Chaz 2.0 Apr 3 '12 at 13:08
Now for your question about why it's not $0$ or negative or whatever, look at the very last formula: $$R = \sqrt[3]{\dfrac{V}{\pi}}$$. If $V = 0$, then $R = 0$. If $V < 0$ (not that this makes sense in the physical context), then $R < 0$. But you can't just say that $V = 0$. You made that up. – The Chaz 2.0 Apr 3 '12 at 13:11
@TheChaz I think the shouting/emphasis is more likely to come off as rude than emphatic. – Austin Mohr Apr 3 '12 at 17:20

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.

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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 $r=\sqrt[3]\frac{V}{\pi}$.

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