I'm trying to assist a student with the following optimization problem. The instructions say to use the partial derivative test which I'm having trouble with. I can solve the problem with Lagrange multipliers just fine, but the student doesn't know that method yet.
A rectangular metal tank with an open top is to hold $256$ cubic feet of liquid. What are the dimensions of the tank that require the least material to build?
Let $A(x,y,z)=xy+2xz+2yz$. Then we want to find $\min \left\{A(x,y,z) \mid xyz=256\right\}$. First I find the critical points,
$$\begin{cases}\frac{\partial A}{\partial x} = y + 2z = 0 \\ \frac{\partial A}{\partial y} = x + 2z = 0 \\ \frac{\partial A}{\partial z} = 2x + 2y = 0\end{cases} \implies x=y=-2z \implies 4z^3 = 256$$
so that $z=4$, but $x$ and $y$ both turn out to be negative. Is there something I am missing, or is this problem flawed in some fundamental way?
Best Answer
$A(x,y,z)=xy+2xz+2yz ~$ itself has no minimum.
So you should apply the constraint and then take the partial derivatives.
$ \displaystyle A = xy + 2x \cdot \frac{256}{xy} + 2y \cdot \frac{256}{xy} = xy + \frac{512}{y} + \frac{512}{x}$
$ \displaystyle \frac{\partial A}{\partial x} = y - \frac{512}{x^2} = 0$
$ \displaystyle \frac{\partial A}{\partial y} = x - \frac{512}{y^2} = 0$
So, $x^2y = xy^2 = 512 \implies x = y = 8$ and that leads to $z = 4$.