# Maxima,Minima and saddle points for functions of two variables

Although I have understood the problem, I am having trouble setting up the function so that i can take the partial derivative and set it equal to zero to determine the critical point. Please help me on how to set up the equation such that it can be differentiated partially based on the below mentioned problem. That's all I am looking for.

An open rectangular container is to have a volume of $32m^3$. Determine the dimensions and the total surface area such that the total surface area is a minimum.

I tried setting it up as $v = 32m^3$,but that seem to not work. I am looking for a step by step solution.

Thank You

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If the dimensions of the container are $x$, $y$ and $z$ (in m), the problem reduces to minimize $S=xy+2(yz+zx)$ subject to $V=xyz=32$ ($x, y, z \geq 0$). (Can you figure out why the formula $S = xy+2(yz+zx)$ holds for the surface area?) You could either proceed via Lagrange multipliers, or you can eliminate $z$ from $S$ using $z = 32/(xy)$. Does this hint help you? –  Srivatsan Aug 31 '11 at 13:04
Srivatsan:I have no idea about Lagrange Multipliers.All I'am Looking for is a specific function of which i can take the partial derivative.I can answer further questions only after taking a partial derivative's of respective variables. –  alok Aug 31 '11 at 13:10
Then take the part of Srivatsan's comment following the word "or". –  Gerry Myerson Aug 31 '11 at 13:15
Srivatsan and Gerry:Ah..I could partially understand.Ok i think you are asking me to substitute z=32/xy in the surface area formula which i found in some text. as S = xy + 2yz + 2xz.How come I don't see a 2 just before xy in the surface area formula ?The rest has got the number 2. –  alok Aug 31 '11 at 13:17
alok: Please do not use [homework] as a single tag, it is only to hint that the problem is a homework exercise. Could you add another tag relating this question to a more concrete mathematical topic? –  Asaf Karagila Aug 31 '11 at 14:35

Suppose that the dimensions of the open container are $x$, $y$ and $z$ (in meters, I'll ignore units henceforth) with $z$ being the height. Then first, as a simple exercise, convince yourself that the volume of the container is $V = xyz$ and the surface area $S = xy + 2(yz+zx)$.

To see why the surface area formula is true, imagine painting the container from the outside, and list the faces that you would need to paint. In particular, how many faces would you need to paint? (Remember that the container is open.)

Going back to the problem, one can rewrite the given volume condition to be $z = \frac{V}{xy} = \frac{32}{xy}$. Plugging this in the formula for the surface area, we get $S(x,y) = xy + 2(y \cdot \frac{32}{xy}+ x \cdot \frac{32}{xy}) = \ldots$ (I'll skip simplifying this further).

Thus the given problem reduces to finding the minimum value of $S(x,y)$ subject to the constraints $x > 0$ and $y > 0$. (Note that the volume condition is not explicitly relevant to us anymore.) Can you take it from here?

Note. The formula for $S$ I have written down considers only the external surface. But since the container is open, it might be more sensible to speak of the total, i.e., external+internal, surface area. In this case, one would need to multiply the formula for $S$ by $2$. (How would this affect the answers?)

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Yes.I can take it from here.However I'am still unclear about the asymmetry part –  alok Aug 31 '11 at 13:23
@alok I have added a bit more explanation. Hopefully that helps. –  Srivatsan Aug 31 '11 at 13:27