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Could somebody please shed some light on this problem?

Let $x,y \in \mathbb R$, we wish to maximize $f(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2}$ by finding suitable values of $x,y$.

Setting $\partial f\over \partial x$ and $\partial f \over \partial y$ as $0$ gives

$x(x^2-3y^2)=0=y(y^2-3x^2)$ but these give $x=y=0$ which is not acceptable! Any ideas?

Thank you.

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So you've proven that the only place the $\nabla f$ could possibly be $0$ is where the function is not well defined. What does that tell you? (Hint: don't think too hard!) –  Neal May 10 '12 at 14:26
    
Your function is clearly unbounded, why are you expecting a different answer? If you are expecting finite answer, are you sure your domain is not bounded in anyway? –  TenaliRaman May 10 '12 at 14:30
    
Thanks Neal and TenaluRaman! –  keane May 10 '12 at 15:51

1 Answer 1

up vote 3 down vote accepted

The function doesn't attain a maximum. Assume that $(x,y)$ was a maximum, then you have $$\frac{x^2-y^2}{(x^2+y^2)^2}\leq \frac{x^2}{(x^2)^2}=\frac{1}{x^2},$$ which is also a value of $f$. Hence $y=0$. But then $f(x,0)\to\infty$ as $x\to 0$.

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