Showing that $\forall (x,y) \in (\mathbb{R}_+)^2$, $x^2 + y^2 \leq 4 e^{x+y-2}$

Question

Let $F = (\mathbb{R}_+ )^2$ and $f: F \rightarrow \mathbb{R}$ defined as $f(x,y) = (x^2 + y^2) e^{-x-y}$.
1) Show that $f$ is bounded on $F$
2) Find the maximum of $f$ on $F$ and deduct that $$\forall (x,y) \in (\mathbb{R}_+)^2, x^2 + y^2 \leq 4e^{x+y-2} $$

I first calculate its partial derivatives:

$$\frac{\partial f}{\partial x} (x,y) = e^{-x-y}(2x - x^2 -y^2), \frac{\partial f}{\partial x} (x,y) = e^{-x-y}(2y - y^2 -x^2)$$

We have $\frac{\partial f}{\partial x} (x,y) = \frac{\partial f}{\partial y} (x,y) = 0$ whenever $(x,y)=(0,0)$ or $(x,y)=(1,1)$

I then calculated the Hessian matrix, I got: $$H(0,0)=\begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}, H(1,1) = \begin{bmatrix} 0 &-2e^{-2} \\ -2e^{-2}& 0 \end{bmatrix} $$

Thus I can deduce that $f$ reaches it's global maximum at $(1,1)$ thus is bounded on $F$.

Obvioously, I want to say that the maximum of $f$ is $f(1,1) = 2e^{-2}$. And then I can conclude that for any $x,y \in F$ we have $f(x,y) \leq 2e^{-2}$, thus $x^2 + y^2 \leq 2e^{-2 + x + y}$. But apparently I need to find a $4$ instead of a $2$ in the inequality. Where is my error?

Answer 1

$a)$ We have: $ f(x,y) \le (x+y)^2\cdot e^{-(x+y)}$. So let $t = x+y > 0 \implies f(x,y) \le g(t) = t^2e^{-t}$. Observe that $g'(t) = e^{-t}(2t-t^2) = 0 \iff t = 0,2$. And $g'(t) > 0$ if $0 < t < 2$, and $g'(t) \le 0$ if $t \ge 2$. Both cases give $g(t) \le g(2) = 4e^{-2}$,and this shows that $f(x,y) \le 4e^{-2}$ which is an upperbound for $f(x,y)$.

For finding the maximum of $f(x,y)$, your work is correct since the domain of $f$ is an open subset of $\mathbb{R}^2$, thus the only critical point $(1,1)$ yields a global maximum of $2e^{-2}$ in your work. Note that $\mathbb{R_{+}}^2 = \{ (x,y): x > 0, y > 0\}$.

$b)$ $f(x,y) \le 4e^{-2}\implies (x^2+y^2)e^{-x-y} \le 4e^{-2}\implies x^2+y^2 \le 4e^{x+y-2}$ . This explains the $4$ that you mentioned in the post.

Answer 2

Hint: $(1,1)$ isn't the global maximum. In fact, from the Hessian matrix it isn't even a local maximum. For a bounded region, Hessian analysis can only gives minima/maxima on the interior of the region - what happens on the boundary?

Answer 3

You better look at $f$ on the closed set $\bigl({\mathbb R}_{\geq0}\bigr)^2$ and then analyze as well the function $\phi(x):=f(x,0)$ $\>(x\geq0)$.Note that your $H(1,1)$ does not signal a local maximum.