Find Min/Max of f(x,y)=xy with Constraint 4x^2+9y^2=32

snoggerT
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find min/max:

f(x,y)=xy with constraint being 4x^2+9y^2=32

[gradient]f=[lambda]gradient g

The Attempt at a Solution



I thought I understood the Lagrange problems, but I can't seem to get the minimum right on the last few problems. I get x=+/-2 and then plug back into find y, then I use my critical points to find my min/max in f(x,y). I got 8/3 for my max on my problem (which is right), but can't get the minimum right. I set it up as such:

f(-2,-4/3)=xy and get +8/3 again, but the answer in the back of the book is -8/3 for the minimum. What am I doing wrong?
 
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You aren't doing anything wrong. But how about f(2,-4/3) or f(-2,4/3)? Nothing in the problem forces x and y to have the same sign.
 
Dick said:
You aren't doing anything wrong. But how about f(2,-4/3) or f(-2,4/3)? Nothing in the problem forces x and y to have the same sign.

- Alright, then that makes sense. The book isn't very good at pointing things out like that.

while on the topic of lagrange...When you get into having 3 variables and 1 constraint, would you set up the problem as [lambda]=x=y=z? If so, how would you solve for the unknowns?
 
Well, you'll have three equations involving x,y,z and lambda coming from the gradient. Then you have the constraint equation in x,y,z. That's four equations in four unknowns. There's no special way to set it up, just derive the equations from the partial derivatives.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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