Maximizing $f(t,x,y) = x^2 e^{- t} + y^2 e^{-2 t}$ subject to constraint $x^2 + y^2 \leq 1$.

Question

Consider the function $$ f(t,x,y) = x^2 e^{- t} + y^2 e^{-2 t} \ . $$ Suppose that $x^2 + y^2 \leq 1$ and $t \geq 0$.

How would one maximize the function $f$ over this domain? By playing around with the function I am pretty sure that the maximum should occur at $t=0$ and $x^2 + y^2 = 1$, but how would you prove this?

EDIT: I would also be very interested in finding a proof for which $t$ and $x$ and $y$ this maximum is obtained. Is it really true that the maximum $1$ is attained on the set $\{ (t,x,y) | t=0 \mathrm{\ and\ } x^2 +y^2 = 1 \}$?

Answer 1

$f(t,x,y) \leq e^{-t}x^{2}+e^{-t}y^{2} \leq e^{-t}$. The maximum value of $f$ for fixed $t$ is $e^{-t}$ since this is attained when $x=0$ and $y=1$. The maximum value over $t,x$ and $y$ is $1$ which is attained at $(0,1,0)$.

Answer 2

Well, since $t\ge 0$, $e^{-2t}\le e^{-t}\le 1$, and $x^2,y^2\ge 0$ so $$x^2e^{-t}+y^2e^{-2t}\le x^2+y^2\le 1$$ And $1$ is attained: $f(0,x,y)=1$ on the circle $x^2+y^2=1$.