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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tiburtius eNotes educator | Certified Educator

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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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