Find minimum value of multivariable-function A tent with 2 rectangle shaped sides (no floor) and 2 isosceles triangles shaped gables with the volume $V$ is to be constructed. Determine the height so that the minimum amount of cloth is needed.
The tent is a prism with isosceles triangle bases. Let the height of the triangle (i.e. the height of the tent) be denoted $x$, the base $2y$ and the length of the tent $L$. Then the volume is $$V(x,y,L)=xyL$$
and the area of the tent will be $$A(x,y,L) = 2(xy+L\sqrt{x^2+y^2}).$$
Since A is a continuous function on a compact set (or can this actually be said, since V is not a boundary but a function of the variables?), there will be a minimum and maximum value. These are found when $grad \, V \, || \, grad \, A$. Since $$ grad \, V = (yL, xL, xy)$$ and $$grad \, A =2(y + \frac{xL}{\sqrt{x^2+y^2}}, x + \frac{yL}{\sqrt{x^2+y^2}}, \sqrt{x^2+y^2})$$
we must find $\lambda$ such that
\begin{cases} y+\frac{xL}{\sqrt{x^2+y^2}} = \lambda yL \\ x+ \frac{yL}{\sqrt{x^2+y^2}} = \lambda xL \\ \sqrt{x^2+y^2} = \lambda xy \end{cases}
I have no idea how to solve this or if this even would be the correct approach. Any help is appreciated.
 A: You can continue with your approach. Adding the first two equations together:
$$(x+y)(1+\frac{L}{\sqrt{x^2+y^2}}-\lambda L)=0$$
This gives you either $x=-y$ or $\lambda=\frac{1}{L}+\frac{1}{\sqrt{x^2+y^2}}$. Since $x=-y$ is not possible, we will continue with the other one. Plugging that into equation 1 or 2, you will get $x=y$.
Now if you replace all $x$ with $y$, you can get a relationship between $x$ and $L$. Remember that $V=xyL$ and $V$ is a constant. With that you can find $x,y$ and $L$.
I don't think your argument with bounded region holds for the minimum because the region is not bounded. It is kind of a hyperbaloid. I think you can either draw a rough picture to see there is a minimum, or use second derivative test on the $2D$ function
$$A(x,y,L) = 2(xy+L\sqrt{x^2+y^2})=2(xy+\frac{\sqrt{x^2+y^2}}{xy})$$
to test the minimum.
A: Because volume V is fixed or constrained,
L = V/xy
substitute above eq into Area eq,
A(x,y) = 2(xy+V/xy*(x^2+y^2)^1/2)
this is minimization problem wrt x and y for area with fixed volume constraint. x, y within (0, inf)
because there are 2 indep variable, not only are you solving for tent height, but also tent base.
i made surface plot for Volume=5, to graphically see where/if the min is

based on plot, there do not seem to single min point, but may have multiple min point.
you can analytically for min point taking grad(A) wrt x and y and then solve for when grad(A) = 0 vector.  you may find no single min point exist
