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
 A: 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?) 
