Question

Multivariate Calc question:

Find all relative extrema of x^2y^2 subject to the constraint 4x^2 + y^2 = 8

Thanks in advance!

Answer 1

I think:
Critical points in the interior are when both partials=0 (or one doesn't exist). This means that the only interior critical point is (0,0)

Answer 2

Or is this a lagrange multipliers question?

Answer 3

If it is Lagrange, would the system be:

\[<2y ^{2} x, 2x ^{2}y> = \lambda<8x, 2y>\]

?

Answer 4

Yeah, I'm pretty lost on this one.

Answer 5

hm... (0,0) is not within the constraint 4x^2 + y^2 = 8, so it's out of bounds
I do think it's that kind of problem though, not lagrange

Answer 6

it is an equals sign in the constraint, not less-than or equal-to, right?

Answer 7

Correct.

Answer 8

That's what made me think Lagrange after I realized that (0,0) isn't in the domain.

Answer 9

Ooh. We know x and y can't both be 0 because that isn't in the domain, so you can cancel terms in the Lagrange equation, yielding:
2y^2 = 8 and
2x^2 = 2
along with 4x^2 + y^2 = 8

Answer 10

y = 2 and x = 1 according to that, so z=4

Answer 11

or -1 and -2, which also yields 4

Answer 12

yes, I came to the same conclusion by doing a simple substitution
lagrange is unnecessary I guess

Answer 13

Although that doesn't yield any minima... which I can't imagine is true given those equations

Answer 14

I think you were right actually, it is a Lagrange problem...
in which case I can't really help since I'm not so good at it
I'll have to read up on it, but hopefully I can answer it next time

Answer 15

Great, well thanks anyway. I really appreciate the help!