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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.

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Please add a tag in addition to [homework] that indicates the mathematical area in question. Here, [multivariable-calculus] might be one possibility. Having only the [homework] tag is not helpful for finding this question later. –  Arturo Magidin Jun 21 '11 at 19:51
    
What about $(x,y)=(\pm1,0)$ and $(x,y)=(0,\pm\frac12)$... –  Did Jun 21 '11 at 19:56
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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
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+1 for showing your work. –  Eric Naslund Jun 21 '11 at 19:59
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1 Answer

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,

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Thanks, I forgot the part where $x=0$ satisfies the first equation. –  Tass Jun 21 '11 at 20:03
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