Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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 $x\to 0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.