How Do You Use Lagrange Multipliers to Maximize Box Volume in an Ellipsoid?

[ryan8888]

1. Problem Statement:

Use Lagrange multipliers to find the volume of the largest box with faces parallel to the coordinate system that can be inscribed in the ellipsoid: 6x² + y² + 3z² = 2

2. Homework Equations :

f(x,y,z) = [tex]\lambda[/tex]g(x,y,z)

3. Attempt at a solution

f(x,y,z) is the box of dimensions xyz
g(x,y,x) is the constraint: 6x² + y² + 3z² = 2

Therefore:

f_x = yz
f_y = xz
f_z = xy

g_x = 12x
g_y = 2y
g_z = 6z

<f_x, f_y, f_z> = [tex]\lambda[/tex]<g_x, g_y, g_z> = <[tex]\lambda[/tex]g_x, [tex]\lambda[/tex]g_y, [tex]\lambda[/tex]g_z>

Which gives us these equations:

yz = [tex]\lambda[/tex]12x (1)
xz = [tex]\lambda[/tex]2y (2)
xy = [tex]\lambda[/tex]6z (3)
6x² + y² + 3z² = 2 (4)

Multiplying Eq 1 by x, Eq 2 by y and Eq 3 by z we get:

xyz = x[tex]\lambda[/tex]12x or [tex]\lambda[/tex]12x²
xyz = y[tex]\lambda[/tex]2y or [tex]\lambda[/tex]2y²
xyz = z[tex]\lambda[/tex]6z or [tex]\lambda[/tex]6z²

Now because [tex]\lambda[/tex] [tex]\neq[/tex] 0 (this would give the sides of the boxes as xz=yz=xy=0) we can divide [tex]\lambda[/tex] out:

and we have 12x² = 2y² = 6z²

This is the point where I am running into trouble. I need to solve the system of equations and I also know that the solution is staring me in the face I just can't seen to be able to figure it out.

Any help is greatly appreciated!

 

[AlephZero]

6x² + y² + 3z² = 2 (4)

and we have 12x² = 2y² = 6z²

which means that 6x² = y² = 3z² = t, say

Substitute into (4) to find t.

 

[HallsofIvy]

Well explained, AlphaZero.

But, because that last part depended on that very nice form for the constraint, you could have done this:
[tex]12x^2= 2y^2= 6z^2[/tex]
dividing through by 2 gives, as AlphaZero said
[tex]6x^2= y^2= 3z^2[/tex]
so that [itex]x= \pm y/\sqrt{6}[/itex] and [itex]z= \pm y/\sqrt{3}[/itex]
and you can replace x and z in whatever constraint you have to get a single equation in y.

Here, of course, you just get [itex]y^2+ y^2+ y^2= 3y^2= 2[/itex] and can solve for y. When finding x and z, don't forget the [itex]\pm[/itex]

 

[ryan8888]

which means that 6x² = y² = 3z² = t, say

Substitute into (4) to find t.

Please forgive me but I'm not following what you are saying here. I just don't see how 6x² = y² = 3z² = t fits into the equation for the constraint.

 

[HallsofIvy]

He is saying that the common value in the equation 6x= y= 3z can be set equal to t: 6x= y= 3x= t. Now, the constraint equation just happens to be 6x+ y+ 3z= 2. Replacing each of those with t you have t+ t+ t= 3t= 2.

 

[ryan8888]

He is saying that the common value in the equation 6x= y= 3z can be set equal to t: 6x= y= 3x= t. Now, the constraint equation just happens to be 6x+ y+ 3z= 2. Replacing each of those with t you have t+ t+ t= 3t= 2.

I can't believe I didn't catch the dividing out the common!

THanks guys I got it now!