# Minimize area of triangular prism and find the height, when the volume is V

1. Aug 13, 2011

### Inertigratus

1. The problem statement, all variables and given/known data
The triangular prism looks like this: http://www.korthalsaltes.com/photo/triangular_prism.jpg
The triangles are "isosceles" triangles though, with the base facing down.
However, this one is open on the bottom.
They want me to find the height of the prism, when the area is as small as possible and the volume is "V".

I put the bottom of the prism at $z = 0$, and the middle part at $x = y = 0$ meaning that one quarter of the bottom is in each quadrant in the xy-plane.

2. Relevant equations
Area $= 2(2xh) + yz = 4xh + 2yz$
$h = \sqrt{y^2 + z^2}$
$f(x, y, z) = 4xh + 2yz$
$g(x, y, z) = 2xyz$
$g(x, y, z) = V$ (a constant)
$\frac{dh}{dy} = -\frac{y}{h}$
$\frac{dh}{dz} = -\frac{z}{h}$
$\frac{df}{dx} = 4h$
$\frac{df}{dy} = 2z - \frac{4xy}{h}$
$\frac{df}{dz} = 2y - \frac{4xz}{h}$

3. The attempt at a solution
$\nabla f(x, y, z) = 0$ gives:
$h = 2z$ (from the partial derivative w.r.t. z)
$h^2 = 4xy$ (from the partial derivative w.r.t. y)
Plugging this into $g(x, y, z) = \frac{h^3}{4} = V$
Obviously the height is maximal when $y = 0$ and therefor we can change say that $h^3 = z^3$.
However, the answer is supposed to be the qubic root of $\frac{V}{\sqrt{2}}$.
Any ideas what might be wrong?

Last edited: Aug 14, 2011
2. Aug 13, 2011

### e(ho0n3

By area, do you mean the surface area of triangular prism or the area of the triangle?

3. Aug 14, 2011

### Inertigratus

Yes, the surface area. I have to find the height of the tent (triangular prism) when the use of material is minimized, meaning as small a surface area as possible.
Note, the tent has no bottom side, so it's two triangles and two rectangles.

(just noticed that one of the partial derivatives were wrong, will see if I get the correct answer after fixing it)

Yay, solved it... sorry if I confused someone, I messed up the derivative w.r.t z.

Last edited: Aug 14, 2011