max of a function in an interval

Question

I have the following function :

$\begin{equation} \displaystyle f(x,y) = \frac{(xy+x+y)}{x^2y^2}\end{equation}$

With $x,y \in [1.75,2.0]$, I already know the max value on that interval is given by $x=y=1.75$ (done with Octave) but how can I prove this?

Answer 1

Note that your function $$\begin{equation} \displaystyle f(x,y) = \frac{(xy+x+y)}{x^2y^2}\end{equation}$$is symmetrical with respect to y=x.

Therefore your minimum and maximum on the box $[1.75,2]\times [1.75, 2]$ is on the line $y=x.$

The function $f(x,x) = \frac {x+2}{x^3}$ is a decreasing function on $[1.75,2].$

Thus the maximum value of " .6997..." is attained at $x=1.75$

Answer 2

Hint: $\;\displaystyle f(x,y) = \frac{1}{xy}+\frac{1}{xy^2}+\frac{1}{x^2y}\,$ is strictly decreasing in each of $\,x,y\,$ on $\mathbb{R}^+\,$.

Answer 3

That is, if $x\geq a$ and $y\geq b$ for $a,b,x,y\in \mathbb{R}^+$ then $f(x,y)\leq f(a,b)$ so the coordinate "closest" to the origin is the maximum value.