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

I got to find the extrema of $f(x,y) = (4x^2+y^2)e^{-x^2-4y^2}$ as usual, first derviative and find roots ($e$ already cleaned out): $$8x+(4x^2+y^2)(-2x)=0$$ $$2y+(4x^2+y^2)(-8y)=0$$ But I can't solve that equation for any other than $(x,y)=(0/0)$. After that, I get $$4-4x^2-y^2=0$$ $$1-16x^2-4y^2=0$$ but I can't squeeze any roots from that, there should be some more.

share|cite|improve this question
What about $(x,y)=(\pm1,0)$ and $(x,y)=(0,\pm\frac12)$... – Did Jun 21 '11 at 19:56
From the first equation, you get $x=0$ or $4x^2+y^2=4$. If $x=0$, then from the second equation you get $2y-8y^3=0$, or $y(1-4y^2)=0$; this gives $x=y=0$ and $x=0, y=\pm 1/2$ as possible solutions. If $x\neq 0$, then you get $4x^2+y^2=4$; plugging that into the second equation you get $y=0$, hence $4x^2=4$, so $x=\pm1$, $y=0$ are also solutions. In summary, you get $(0,0)$, $(\pm 1,0)$, $(0,\pm\frac{1}{2})$. – Arturo Magidin Jun 21 '11 at 19:58
+1 for showing your work. – Eric Naslund Jun 21 '11 at 19:59
up vote 3 down vote accepted

Hint: Above you noted that $(0,0)$ is a solution, and then considered the case where $y\neq 0$ and $x\neq 0$. This case as you remarked has no solutions since $4x^2+y^2$ cannot take on two different values. But is this all the cases?

What about when $x=0$, and $y\neq 0$? Then $8x+(4x^2+y^2)(-2x)=0$ is satisfied, and the second equation after division by $y$ becomes $$\frac{1}{4}=y^2$$ yielding the two additional pairs $(0,\frac{1}{2}),\ (0,-\frac{1}{2})$.

Similarly, what happens when $y=0$ and $x\neq 0$? We should get $2$ additional solutions from this case.

Hope that helps,

share|cite|improve this answer
Thanks, I forgot the part where $x=0$ satisfies the first equation. – Reactormonk Jun 21 '11 at 20:03

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.