# Lagrangian Multipliers

• Oct 20th 2009, 11:07 PM
Aryth
Lagrangian Multipliers
A box has three of its faces in the coordinate planes and one vertex on the plane 2x + 3y + 4z = 6. Find the maximum volume for the box.

Any help would be greatly appreciated.
• Oct 21st 2009, 08:36 AM
Opalg
If the sides of the box in the directions of the three coordinate axes have lengths x, y and z, then the volume of the box is xyz. The only vertex of the box that does not lie in one of the coordinate planes is the one at (x,y,z), and that has to satisfy 2x + 3y + 4z = 6. So you should use the method of Lagrange multipliers to maximise xyz subject to the constraint 2x + 3y + 4z = 6.
• Oct 21st 2009, 09:32 AM
Aryth
Ok... So I gave it a try... I set up the functions as so:

$V(x,y,z) = xyz$

$\phi(x,y,z) = 2x + 3y + 4z - 6$

$F(x,y,z) = V(x,y,z) + \lambda \phi(x,y,z)$

$F(x,y,z) = xyz + \lambda{2x} + \lambda{3y} + \lambda{4z} - \lambda{6}$

Taking the partial derivatives:

$\frac{\partial{F}}{\partial{x}} = yz + 2\lambda = 0$

$\frac{\partial{F}}{\partial{y}} = xz + 3\lambda = 0$

$\frac{\partial{F}}{\partial{z}} = xy + 4\lambda = 0$

This gives me:

$yz = -2\lambda$
$xz = -3\lambda$
$xy = -4\lambda$
$2x + 3y + 4z = 6$

First, I noticed that:

$xy = 2yz$
$z = \frac{1}{2}x$

Then, I noticed that:

$xz = \frac{3}{2}yz$
$y = \frac{2}{3}x$

Now, I plugged those to into the constraint function:

$2x + 3\left(\frac{2}{3}x\right) + 4\left(\frac{1}{2}x\right) = 6$

$2x + 2x + 2x = 6$
$x = 1$

Which gave me:

$y = \frac{2}{3}$
$z = \frac{1}{2}$

Did I do it right? I'm not too familiar with this method... I missed the class that covered it and I'm trying to understand it. (Headbang)