Min and Max of $ f(x,y)=\frac{x-y}{a-x-y}$

Question

I'd like to find max and min of $$ f(x,y)=\frac{x-y}{a-x-y}$$ where $0\le x<y\le a/2$. Any one can suggest? Thank you

Answer 1

Since $f_{x}=\frac{a-2y}{(a-x-y)^2}$ and $f_{y}=\frac{2x-a}{(a-x-y)^2}$, f does not have any critical points in the interior of the closed region bounded by the lines $x=0$, $y=x$, and $y=\frac{a}{2}$.

On the boundary of this region,

$f(x,y)=0$ on the line $y=x$;

$f(x,y)=-1$ on the line $y=\frac{a}{2}$;

and on the line $x=0$, $h(y)=f(0,y)=\frac{y}{y-a}$ is decreasing since $h^{\prime}(y)=-\frac{a}{(y-a)^2}$;

$\;\;\;$so $h(0)=0$ is its maximum, and $h(\frac{a}{2})=-1$ is its minimum.

Since the region being considered is the closed region above with the line segment lying on $y=x$ deleted, f does not have a maximum (but has values arbitrarily close to 0) and has a minimum of -1.

Answer 2

Thats what Wolfram Alpha says: http://www.wolframalpha.com/input/?i=maximize+%28x-y%29%2F%28a-x-y%29+on+0%3C%3Dx%3Cy%3C%3Da%2F2

So the maximum is basically where $$x-y\rightarrow 0$$

Answer 3

Since $0\le x<y\le a/2$, we have that $x-y<0$ and $a-x-y>0$. Hence $f(x,y)<0$ everywhere on the given domain. However, for $x+y<a$, we have $\lim\limits_{x\rightarrow y}f(x,y)=0$. This means that $\sup f(x,y)=0$, but this supremum is never achieved.

Answer 4

If x < y <= a/2, then x - y is always negative but a - x - y is always positive. That means f(x,y) must be a negative number.

So the best we can do is 0 - epsilon. Notice that the closer you get to x = y, the closer the numerator approaches zero while the denominator is bounded below by a. So the max of this function does not exist (its supremum is zero, but any number below zero is possible and there is no smallest negative number).