- #1

- 128

- 0

## Homework Statement

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 [itex]z = 0[/itex], and the middle part at [itex]x = y = 0[/itex] meaning that one quarter of the bottom is in each quadrant in the xy-plane.

## Homework Equations

Area [itex]= 2(2xh) + yz = 4xh + 2yz[/itex]

[itex]h = \sqrt{y^2 + z^2}[/itex]

[itex]f(x, y, z) = 4xh + 2yz[/itex]

[itex]g(x, y, z) = 2xyz[/itex]

[itex]g(x, y, z) = V[/itex] (a constant)

[itex]\frac{dh}{dy} = -\frac{y}{h}[/itex]

[itex]\frac{dh}{dz} = -\frac{z}{h}[/itex]

[itex]\frac{df}{dx} = 4h[/itex]

[itex]\frac{df}{dy} = 2z - \frac{4xy}{h}[/itex]

[itex]\frac{df}{dz} = 2y - \frac{4xz}{h}[/itex]

## The Attempt at a Solution

[itex]\nabla f(x, y, z) = 0[/itex] gives:

[itex]h = 2z[/itex] (from the partial derivative w.r.t. z)

[itex]h^2 = 4xy[/itex] (from the partial derivative w.r.t. y)

Plugging this into [itex]g(x, y, z) = \frac{h^3}{4} = V[/itex]

Obviously the height is maximal when [itex]y = 0[/itex] and therefor we can change say that [itex]h^3 = z^3[/itex].

However, the answer is supposed to be the qubic root of [itex]\frac{V}{\sqrt{2}}[/itex].

Any ideas what might be wrong?

Last edited: