Calculus 3, find absolute max/min difficult problem?

Question

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

D is the disk x^2 + y^2 < or equal 4.

Answer 1

This problem can be greatly simplified by converting from Cartesian to polar coordinates.

Specifically, use the change of variables

x = r cos θ
y = r sin θ;

r = sqrt( x² + y² )

That transforms f(x,y) into

f(r,θ) = exp(-r²)(r² + [r sin θ]² ),

and the domain in question to

r ε (Real) | 0 ≤ r ≤ 2.

It may be more convenient to distribute the exponential part for clarity, so that

f(r,θ) = r² exp(-r²) + r² sin² θ exp(-r²).

Since both terms contain r², and since sin² θ is always positive, it is relatively simple to show that the absolute minimum occurs at r = 0.

i.e. f(0,θ) = 0 ≤ f(r,θ)

It is not clear, by inspection, whether any *local* maxima occur within the given domain. That means that you have to solve

∂f / ∂r = ∂f / ∂θ = 0,

where the solutions are maxima if

∂²f / ∂r² < 0
∂²f / ∂θ² < 0

(Note: I've got to go to bed now, so I'll let you find the partial derivatives, and solve yourself. I will say, that if no local extrema occur within the domain, then the absolute max lies at

(r,θ) = (2,π/2), and (2, 3π/2)........)

Good luck,
~W.O.M.B.A.T.