Question

Find the absolute maximum of f(x,y) = e^(-x^2-y^2)(x^2+2y^2) on x^2 + y^2 < 2

Answer 1

`f(x,y)=e^(-x^2-y^2)(x^2+y^2)`

Notice that converting to polar coordinates, setting

`x=r\cos\theta`

`y=r\sin\theta`

`\implies r^2=x^2+y^2`

allows us to consider
`f(x,y)` as a function of one variable; let's call it
`F(r)`, where


`f(x,y)\equiv F(r)=re^(-r)`

Then


`F'(r)=e^(-r)(1-r)=0\implies r=1`

We have
`F'(r)>0` for
`r<1`, and
`F'(r)<0` for
`r>1`, which means
`F` is increasing, then decreasing as
`r` exceeds 1. This suggests that extrema occur for
`f(x,y)` wherever
`r^2=x^2+y^2=1`, i.e. along the intersection of the cylinder
`x^2+y^2=1` and
`f(x,y)`.

Computing the second derivative of
`F(r)` and setting equal to 0 gives


`F''(r)=-e^(-r)(2-r)=0\implies r=2`

as a possible point of inflection. We have
`F''(r)<0` for
`r<2`, and namely when
`r=1`, which means
`F(r)` is concave downward around this point. This confirms that
`r=1` is a site of a maximum. Along this path, we have a maximum value of
`F(1)=e^(-1)\approx0.368`.

Next, to check for possible extrema along the border, we can parameterize
`f(x,y)` by
`x=\sqrt2\cos t` and
`y=\sqrt2\sin t`, so that


`x^2+y^2=(\sqrt2\cos t)^2+(\sqrt2\sin t)^2=2`

and we can think of
`f(x,y)` as a function a single variable,
`F(t)`, where


`F(t)=2e^(-2)\approx0.271`

In other words,
`f(x,y)` is constant along its boundary
`x^2+y^2=2`, and this is smaller than the maximum we found before.

So to recap, the maximum value of
`f(x,y)` is
`\frac1e\approx0.368`, which is attained along the surface above the circle
`x^2+y^2=1` in the
`x-y` plane.