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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.2012-04-03
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    But I don't understand how 0 isn't the answer.2012-04-03
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    Because a cylinder with zero area has zero volume. So unless $V=0$, the answer cannot be $0$.2012-04-03
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    The problem doesn't specify that the volume is > 0.2012-04-03
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    @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...2012-04-03
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    Does the problem ask you to show or state why the solution you find is a global minimum? Have you been shown the second derivative test, what it means for a function to be convex or concave, and the use of the number line of the first derivative to evaluate critical points? Usually, this material is also covered, so that you don't just blindly go solving for roots of the derivative without knowing what result it really gives you.2012-04-03
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    Yes I know about concavity and the first and second derivative tests.2012-04-03
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    Dear Jordan, You are (or at least were) confused about the point of the question. The point is: imagine you want to make cans which hold say $400$ cm${}^3$ of liquid as cheaply as possible (i.e. using as little metal as possible), then how should you make them: tall and thin, short and fat, somewhere in between? The point of the question is to work this out, and not just for $400$ cm${}^3$, but for an arbitrary volume $V$. It's not good telling someone who wants to make and sell $400$ cm${}^3$ volume cans of soup that they should make cans that hold *no* soup. You have to tell them how ...2012-04-04
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    ... they can most cheaply make the cans of the size they want. Regards,2012-04-04

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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 $R$.2012-04-03
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    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!!!**2012-04-03
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    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.2012-04-03
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    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.2012-04-03
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    @TheChaz I think the shouting/emphasis is more likely to come off as rude than emphatic.2012-04-03
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    It is okay because all of this is way, way over my head anyways. I can't really follow this at all. Am I suppose to have all this stuff memorized?2012-04-03
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    I don't understand any of the formulas. I can sort of see why the cylinder one works, because that is a circle area and then times the height. So is it assumed that the area of a circle is always 1 unit high? Also how is the surface area not just the area of a rectangle plus the area of a circle?2012-04-03
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    @Jordan: Dear Jordan, The can is $h$ cm high and its bottom is a circle of radius $R$ cm. The bottom then has area $\pi R^2$. The cylinder part is a rectangle (as you noted), which (imagining it standing vertically, so that we can wrap it around to make a can) has a height of $h$ cm, and a base of $2\pi R$ cm (so that it can wrap completely around the circular bottom of the can). So the area of the rectangle is $2 \pi R \times h$. So the total area of metal in the can is $2 \pi R h + \pi R^2$. Regards,2012-04-04
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    @Jordan: Dear Jordan, It may help you to visualize some cans (or even just to look at some different shaped cans, if you have them sitting in the cupboard), and think about their various dimensions (height, radius) and how these affect the volume of liquid that the can can contain. E.g. if you have a given can, and you push it down but simultaneously make the radius wider, you can keep it enclosing the same volume. Will the new-shaped can need more or less metal? (It depends on the original dimensions of the can, and how much squashing you did, but this is what is happening here.) Regards,2012-04-04
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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 $r=\sqrt[3]\frac{V}{\pi}$.