# Local extremes of: $f(x,y) = (x^2 + 3y^2)e^{-x^2-y^2}$

I am looking to find the local extremes of the following function:

$$f(x,y) = (x^2 + 3y^2)e^{-x^2-y^2}$$

What have I tried so far?

1. Calculate the partial derivatives:

$$\frac{\partial f}{\partial x} = 2x(e^{-x^2-y^2}) + (x^2+3y^2)(e^{-x^2-y^2})(-2x)$$ $$=-2(e^{-x^2-y^2})x(-1+x^2+3y^2)$$

$$\frac{\partial f}{\partial y} = 6y(e^{-x^2-y^2}) + (x^2+3y^2)(e^{-x^2-y^2})(-2y)$$ $$=-2(e^{-x^2-y^2})y(-3+x^2+3y^2)$$

1. Determine the possible stationary points: $$\frac{\partial f}{\partial x}=0 \implies -2(e^{-x^2-y^2})x(-1+x^2+3y^2) = 0$$ $$\implies [x=0] \text{ or } -1+x^2+3y^2 = 0 \implies [x=\sqrt{-3y^2+1}]$$

$$\frac{\partial f}{\partial y} = 0 \implies [y=0] \text{ or } -3+x^2+3y^2 = 0 \implies [y=\pm \sqrt{\frac{-x^2+3}{3}}]$$

So now I have: $$\frac{\partial f}{\partial x}\begin{cases} x = 0 ,\\ \sqrt{-3y^2+1}\end{cases} \text{ and } \frac{\partial f}{\partial y}\begin{cases} y = 0 ,\\ \sqrt{\frac{-x^2+3}{3}}\end{cases}$$

From this I determined the following possible points: $$(0,0) \rightarrow f(0,0) = 0$$ $$(0,1) \rightarrow f(0,1) = 3e^{-1}$$ $$(0,-1) \rightarrow f(0,-1) = -3e^(-1)$$ $$(1,0) \rightarrow f(1,0) = e^{-1}$$ $$(\sqrt{\frac{1}{3}},\sqrt{3}) \rightarrow f(\sqrt{\frac{1}{3}},\sqrt{3}) = \frac{28}{3}e^{-\frac{5}{4}}$$ $$(-\sqrt{\frac{1}{3}},\sqrt{3}) \rightarrow f(-\sqrt{\frac{1}{3}},\sqrt{3}) = \frac{28}{3}e^{-\frac{5}{4}}$$ $$(\sqrt{\frac{1}{3}},-\sqrt{3}) \rightarrow f(\sqrt{\frac{1}{3}},-\sqrt{3}) = \frac{28}{3}e^{-\frac{5}{4}}$$

So far, did I follow the steps correctly? Also, from what I know next is to check the diferencial of the function in each point and those which are zeros are extreme points. Is that correct? And what would follow this?

• Hi Omari , below step no.2 , the points are from $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$ right? And also, for partial derivative use $\partial$ symbol instead of $d$. – Arief Anbiya Nov 3 '17 at 14:42
• @Arief yes. What's the code for that symbol? – Omari Celestine Nov 3 '17 at 14:47
• @Moo do you think you can pinpoint where I went wrong? – Omari Celestine Nov 3 '17 at 14:48
• @Omari use  \partial  . Also, you might miss type below step no.2, you put both are from derivative wrt $x$. – Arief Anbiya Nov 3 '17 at 14:51
• @Arief thanks. I made the changes. – Omari Celestine Nov 3 '17 at 14:55

In 1D (single variable Calculus), a stationary-critical point is a point $c$ such that $f'(c)=0$. At $c$ the gradient of $f$ is zero (there is no change).

In 2D, a stationary-critical point is a point $(a, b)$ such that the gradient in $x$ direction and $y$ direction are both zero (at that same point $(a,b)$, simultaneously, as @Moo said). At that point, no change of value of $f$ in both $x$ and $y$ direction. This means :

$$\frac{\partial f(a,b)}{\partial x } = \frac{\partial f(a,b)}{\partial y } = 0$$

So, in your case, you must find a point $(a, b)$ such that both

$$\frac{ \partial f }{\partial x} = - 2 e^{-x^{2} - y^{2}}x \left( -1 + x^{2} + 3y^{2} \right) = 0$$ $$\frac{ \partial f }{\partial y} = - 2 e^{-x^{2} - y^{2}}y \left( -3 + x^{2} + 3y^{2} \right) = 0$$

which means solving both at once. The system to solve is :

• $- 2 e^{-x^{2} - y^{2}}x \left( -1 + x^{2} + 3y^{2} \right) = 0$
• $- 2 e^{-x^{2} - y^{2}}y \left( -3 + x^{2} + 3y^{2} \right) = 0$

Since $e^{...}$ is never $0$, you may solve these instead :

• $x \left( -1 + x^{2} + 3y^{2} \right) = 0$
• $y \left( -3 + x^{2} + 3y^{2} \right) = 0$

You can continue from here to solve the system.

Hope this helps.