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I have a function

$$f(x,y)=\dfrac{x}{x-y},\;\text{for all}\;x\neq y$$

Where $f(x,y)$ is maximum?

I have calculated $\partial f/\partial x = -y/(x-y)^2$ and $\partial f/\partial y = 1/(x-y)^2$ but what to do next?

drzbir
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    Use the second partials test. http://tutorial.math.lamar.edu/Classes/CalcIII/RelativeExtrema.aspx – Decaf-Math Jan 14 '16 at 20:14
  • Because I think that when $x$ is close to $y$, the ratio is big but the value of $x$ will affect this big ratio. No? – drzbir Jan 14 '16 at 20:20
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    @5.r.a For any $x$, as $y\to x\pm$, $f(x,y)\to \mp\infty$, i.e. the function is unbounded positive/negative as $y$ approaches $x$ from below/above respectively. There is no max. – BrianO Jan 14 '16 at 22:17

3 Answers3

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Consider $$g(y) = f(1, y) = \frac1{1-y}$$ Even if $y\neq x = 1$, we can consider the following limit: $$\lim_{y\to 1^-}\frac1{1 - y} = \infty$$ It should be clear that no maximum exists

jameselmore
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If your domain is $\mathbb{R}^{2}$, then $f$ doesn't have a maximum. You can very easily make $f$ arbitrarily large by choosing $y$ close to $x$ (provided $x \ne 0$). Since $\frac{\partial f}{\partial y}\ne 0$, the only place $f$ could possible have a maximum is on its boundary (eg if $|y-x|>k$ for some $k$ and all $x,y$ in your domain)

preferred_anon
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Remember that $\frac{\partial f}{\partial x}$ is the slope at a point $(x, y)$ strictly considering the x direction only. Similarly, $\frac{\partial f}{\partial y}$ is the slope at a point strictly considering the y direction only.

For us to have a maximum, both slopes on both directions at that maximum should be $0$. So we get a system:

$$0=\frac{-y}{(x - y)^2}, 0 = \frac{1}{(x - y)^2}$$

Clearly, there is no solution for that second equation, so there is no point in which $\frac{\partial f}{\partial y}$ is zero. That means we don't get a flat surface anywhere in the y-direction, so no maximum exists.

Kaynex
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