MVC help - lagrange and related

[shikhan]

Hey guys,

I'm not sure if this really classifies as highly technical but for once, OT has failed to answer and hopefully you guys would indulge me with some help. Sorry about crossposting between here and OT...
I'm trying to review a few things from the book and I can't understand two concepts. The first is maximization of a 3 variable equation with bounds. Here is a sample problem:

Find the volce of the largest recangular box with edges parallel to the axes that can be inscribed in the ellipsoid: (x^2/a^2) +(y^2/b^2)+(z^2/c^2)=1


The second concept is lagrange multipliers. It seems similar to the first but the chapter is not helping me at all.

One of the examples i'm trying to understand is this:

find the maximum and minimum values of the fuction using lagrange multiplies and subject to the given constraints:

f(x,y) = 4x +6y; x^2+y^2=13...


This should be pretty easy for the kinds of people i've seen here.... I hate it when books fail to explain a topic...


[edit] Mods, feel free to lock if this strays too far from the HT "feel" and my apologies in advance...

 

[jaak]

Your problems are essentially one and the same apart from some minor details. Obviously, with the first problem you could solve the equation for the ellipse for one variable, say z. Then plug this into the equation for the volume V = xyz, take the first partial derivatives with respect to x and y and set those to zero to get the values of x and y for an extrema and then backsolve the ellipse equation for z. Boy that's a mouthful. So let's break it down a little. To find an extremum of a function f(x,y,z) we use the fact that the differential of f is zero at such a point, df = 0. Doing the math gives formally

df = (df/dx)*dx + (df/dy)*dy + (df/dz)*dz

Since we assume that (x,y,z) are independent, then dx,dy,dz can be assigned arbitrarily and the only way that df can be zero is if each of the partial derivatives is zero. If I now add a constraint equation, I can no longer let dx,dy,dz vary independently. Let g(x,y,z) = 0 be my constraint. Then by forming dg I have a second equation for dx, dy, dz. So what do I do? As mentioned, I could select say x and y and solve g for z, substitute and maximize f in terms of the remaining two variables. Lagrange multipliers give me a trick to avoid all that work.

Since g(x,y,z) = 0, I can multiply it by any constant and add to f(x,y,z) without altering the value. Let

p(x,y,z) = f(x,y,z) + L1*g(x,y,z)

and compute dp. I now say, hmm let L1 have a value such that the term multiplying dz (could pick dx or dy also) by identically zero. Then since I only have two terms left, and two independent variables, the remaining terms must also be zero. Since I don't know the value of L1 a priori, it is undetermined until I solve for it. I do this at the same time I solve for x, y, z using the 3 partials of p and the constraint equation. Note that if I had 2 constraint equations, I would simply add L2*g2(x,y,z) make the same argument for two of the terms disappearing with appropriate values of L1, L2, and the third going away because the last variable is independent.

Does this help?