# Lagrange Multiplier Fix..

• November 16th 2008, 12:38 PM
roonaldo17
Lagrange Multiplier Fix..
Hi, I'm in the midst of revision for a calculus exam next week. I'm not sure if I tackled this problem correctly.

Find the maximum and minimum distances from the origin for points on the curve described by
3x^2 + 2y^2 + z^2 = 20 and z^2 = 2xy .

I know the constraint eqns defined by g(x,y,z)= 3x^2 + 2y^2 + z^2 - 20 = 0 and h(x,y,z)= z^2 = 2xy
Also I define the function f(x,y,z)= x^2 + y^2 + z^2, which I have to find the minimum and maximum values for.

In short, what I did was to introduce a function F(x,y,z,k,j) = f(x,y,z) - k[g(x,y,z)] - j[h(x,y,z)] and solve for grad(F)=0 ( k and j are constants)

The problem is when I realised 1 of the constants I had used turns out to be 0, is that correct?

PLEASE, would really appreciate some guidance on this: my ans is minimum: (-sqrt(13/8), 0, 0) and max: (sqrt(13/8), 0, 0)
• November 17th 2008, 11:35 AM
Moo
Hello,

Here is what I've done.
I'm finding two extreme values : $10$ and $\frac{20}{3}$ (so the first one is a maximum and the second one a minimum), each of them attained at two points.

Sorry for the writing, I've got a problem with writing on a straight line (Rofl)
I hope it looks correct and helps you.

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Note that your answers for min and max yield the same result for f.
Since it is a symmetric function, f((-sqrt(13/8),0,0))=f((sqrt(13/8),0,0))
• November 17th 2008, 08:55 PM
roonaldo17
k+j=1?
Thanks! I could follow until the part when you consider z is not zero.

Shouldnt k+j=1?

If it's not too much trouble could you please explain more on this part?

Many thanks
:D
• November 17th 2008, 11:49 PM
Moo
Yo,

That's basic algebra !

If ab=0, then either a=0, either b=0.

Here, a=z and b=k+j-1

funny enough, my friend did the exact same mistake ^^
• November 19th 2008, 12:54 AM
roonaldo17
THANKS:)
Thanks for your time spent on this. I misread one for lambda, lol..

The factorisation part was really neat, I would have been defeated by tt equation.

Glad tt I've finally understood this qn.

Cheers.