Question

Find the global maximum and the global minimum of the function \(f(x,y)=e^{-x^2-y^2}(x^2+2y^2)\) on the plate region R consisting of all points for which \(x^2+y^2\leq 4\) by finding the critical points of f(x,y) inside R and on the boundary of R. You can use the fact that e>2

Answer 1

So I think my first order of buisness is to find the \(\nabla f(x,y)\) which I get \[\left\langle2xe^{-x^2-y^2}(1-x^2-2y^2),2ye^{-x^2-y^2}(2-x^2-2y^2)\right\rangle\]

Answer 2

now would I set this equal to \[\left\langle 0,0\right\rangle\]

Answer 3

giving me
\[2xe^{-x^2-y^2}(1-x^2-2y^2)=0\]\[2ye^{-x^2-y^2}(2-x^2-2y^2)=0\]

Answer 4

Is this correct so far? If so I can see that (0,0) would be a critical point. so would I just need to solve \[1-x^2-2y^2=0\] and \[2-x^2-2y^2=0\]

Answer 5

No I am really lost here. No Idea what to do.

Answer 6

well one solution you seem to have overooked is \[x=0,y=0\]

Answer 7

overlooked*

Answer 8

I mentioned that actually. I can't figure out the other ones!

Answer 9

you mean you want to look at the boundary?

Answer 10

I keep getting zero for everything... not sure if that's right

Answer 11

I'm not sure if I'm doing anything correct haha.

Answer 12

I am sort of familiar with these problems, but am a bit rusty
here is an example of a... somewhat similar problem
http://tutorial.math.lamar.edu/Classes/CalcIII/AbsoluteExtrema.aspx
(example 2)
but this is harder and more confusing than the example given

Answer 13

didn't think this is calculus

Answer 14

and what did you think it was panlac?

Answer 15

precal

Answer 16

Is there a guide for the equation tool within the website? I see sometimes the text are huge and there are many things that aren't included in the tools.