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...

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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.