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I'm trying to find the critical points of: $$ f(x,y) = (x^2 + 2y^2)e^{1-x^2-y^2} $$

I applied the product rule and chain rule for each of the partial derivatives to get these two: $$f_x(x,y) = 2x(e^{1-x^2-y^2}) + (x^2 + 2y^2)(-2x)(e^{1-x^2-y^2})$$ $$f_y(x,y) = 4y(e^{1-x^2-y^2}) + (x^2 + 2y^2)(-2y)(e^{1-x^2-y^2})$$

When I set those two to zero I can reduce them to: $$x^2 + 2y^2 = 1$$ $$x^2 + 2y^2 = 2$$

Which I'm pretty sure is right. Trying to solve those two I only get nonsense like $1 = 2$ or the square root of a negative. I looked at the answer and the points are $(1,0), (-1,0), (0,1), (0,-1), (0,0)$

I can see that those work out with the two formulas right above but I can't get there logically. Is there a flaw in my reasoning so far or do I just need to keep working to solve those two partials ? Also if $y$ is undefined for a value of x does that make it critical value of $x$ ?

MadMax
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  • Why did you drop $x=y=0?$ – coffeemath Aug 22 '23 at 11:37
  • "$y$ is undefined for a value of $x$" doesn't make sense. You mean that for a value of $x$, there are no values of $y$ where $(x,y)$ is a critical point? – aschepler Aug 22 '23 at 11:39
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    One way you could check your work is to test one of the answer's critical points and see what happens. For example, $(0,0)$ is a critical point according to your initial $f_x$ and $f_y$, but not according to your reduced equations. So then you know that something went wrong when you took that step. – Teepeemm Aug 22 '23 at 21:01
  • Thanks I left (0,0) out because I thought of it as a gimmie – MadMax Aug 23 '23 at 20:31

3 Answers3

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You cancelled out $x$ in the first equation and $y$ in the second. That leaves with the possibility of $x=0$ and $y=0$ providing critical points. In fact, $x=y=0$ gives the absolute minimum of $f$. Note that $x=0, x^{2}+2y^{2}=2$ and $y=0, x^{2}+2y^{2}=1$ also have solutions.

geetha290krm
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You lost some parts when you reduced your equations. They should read \begin{align*} 2x(1-x^2-2y^2)e^{1-x^2-y^2} &= 0,\\ 2y(2-x^2-2y^2)e^{1-x^2-y^2} &= 0. \end{align*} Sure, the exponential does not matter, but this system of equations has more solutions than yours, e.g., $x = 0$ solves the first equation and then you can solve for $y$ in the second one. You also get the solution $(0,0)$, which apparently the answer also missed.

Klaus
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  • How is this different than what I have above ? What I have is just a manipulation of (1−^2−2^2). If either expression in parentheses goes to zero then the whole thing becomes zero that's why I was solving for it individually. Is that correct ? – MadMax Aug 23 '23 at 23:14
  • Yes, but you missed that $x = 0$ and $y = 0$ also cause the equations to be zero. – Klaus Aug 24 '23 at 08:15
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After eliminating the exponential factor (which is never zero), the system to solve should be $$\begin{cases} x(x^2 + 2y^2 - 1)=0\\ y(x^2 + 2y^2 - 2)=0 \end{cases}$$ From the first one we get $$\begin{cases} x=0\\ y(x^2 + 2y^2 - 2)=0 \end{cases} \cup \begin{cases} x^2 + 2y^2=1\\ y(x^2 + 2y^2 - 2)=0 \end{cases} $$ that is $$\begin{cases} x=0\\ y(y^2 - 1)=0 \end{cases} \cup \begin{cases} x^2 + 2y^2=1\\ y=0 \end{cases}$$ or $$\begin{cases} x=0\\ y(y-1)(y+1)=0 \end{cases} \cup \begin{cases} (x-1)(x+1)=0\\ y=0 \end{cases}$$ Therefore the critical points are $(1,0), (-1,0), (0,1), (0,-1)$ AND $(0,0)$ (which is a global minimum point since $f(0,0)=0$ and $f\geq 0$).

Robert Z
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