Minima and maxima of $f(x,y)=e^{\frac{x}{2+y}}$ on $K=[-1,1]\times[-1,1]$

Question

I am trying to determine the minima and maxima of $f(x,y)=e^{\frac{x}{2+y}}$ on $K=[-1,1]\times[-1,1]$.

$f_x=\frac{e^\frac{x}{2+y}}{2+y}\neq0 \ \forall x,y \in \mathbb{R}$ so there are no extrema on $(-1,1) \times(-1,1)$ but as $f$ is continuous and $K$ is compact $f$ must have extrema on $K \setminus (-1,1)\times(-1,1)$.

So I decided to look at

$f(x,1)=e^\frac{x}{3}$

$f(x,-1)=e^x$

$f(1,y)=e^\frac{1}{2+y}$

$f(-1,y)=e^\frac{-1}{2+y}$

$f(1,1)=e^\frac{1}{3}$

$f(-1,-1)=e^{-1}$

$f(1,-1)=e$

$f(-1,1)=e^\frac{-1}{3}$

Therefore $f$ has a global maximum on $(1,-1)$ with $f(1,-1)=e$ and a global minimum at $(-1,1)$ with $e^{-1}$. Is that correct? And if so, how would a rigorous reasoning look like?

Thanks!

Answer 1

A rigorous reasoning would start by calculating the gradient vector (differentiate with respect to each vector), then calculating when it equals zero on the said domain to deduce the variation of this 2D function $f$. However, what you did is kind of equivalent, but not very rigorous.

Answer 2

Figures always help too, and are just as "rigorous" as calculating by hand certain values such as corner cases:

Answer 3

The exponential function is strictly increasing. The given function $f$ takes its minima and maxima therefore at the same points as the exponent $g(x,y)=x/(2+y)$. On the domain $K$, the function $g$ is strictly increasing with respect to $x$ and strictly decreasing with respect to $y$. Hence, the global maximum occurs when $x$ is as large as possible and $y$ is as small as possible, i.e., in the point $(x,y)=(1,-1)$. Analogously, the minimum occurs at $(x,y)=(-1,1)$. No differential calculus is needed.