Find the relative extrema using the First-partials Test: f(x, y)= e^(-x^2 -y^2) this is what I have figured out so far: u=-x^2 -y^2 => -2x-2y=0 =>x=1 and y=1 or x=-2 and y=-2 or Zx=-2y and Zy=-2x...
Find the relative extrema using the First-partials Test: f(x, y)= e^(-x^2 -y^2)
this is what I have figured out so far: u=-x^2 -y^2 => -2x-2y=0 =>x=1 and y=1 or x=-2 and y=-2 or Zx=-2y and Zy=-2x
I'm just not really sure how to work this one, with the e factoring in.
any help would be appreciated
Thanks
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What you need to do is to solve the following system:
`(del f)/(del x)(x,y)=0`
`(del f)/(del x)(x,y)=0`
So let's derivate function `f(x,y)=e^(-x^2-y^2)=e^(-x^2)e^(-y ^2)`.
`(del f)/(del x)(x,y)=e^(-x^2)e^(-y^2)(-2x)=0`
And since exponential equation can never be 0 (`e^(-x^2)e^(-y^2)ne0` ), it follows that
`-2x=0 => x=0`
Similary you get the other result:
` ` `(del f)/(del x)(x,y)=e^(-x^2)e^(-y^2)(-2y)=0`
Again exponential function is never equal to 0 hence
`-2y=0 => y=0`
So local extrema is: `(x,y)=(0,0)` and `f(0,0)=1`
In this case that is also global extrema, global maximum to be precise.
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