Another max value on surface (no boundaries)

[Addez123]

Homework Statement

$$f(x,y) = (x^2 +2y^2)e^{-x^2-y^2}$$
where
$$x, y \in R$$

Relevant Equations

Math

I've double-checked my equations and can't find what's wrong.
First I calculate the partials:
$$f_x = (2x - 2x^3 -4xy^2)e^{-x^2, -y^2}$$
$$f_y = (-2yx^2 + 4y - 4y^3)e^{-x^2, -y^2}$$

By setting f_x = 0 I get:
$$x^2 = 1 - 2y^2$$
Then I calculate f_y = 0
$$-2yx^2 + 4y - 4y^3 = 0$$
I plug in the results from f_x into my f_y equation and get
$$-2y(1 - 2y^2) + 4y - 4y^3 = 0 $$
$$y = 0 => x = +-1$$
So my critical points are (1, 0) and (-1, 0)
$$f(1,0) = e^-1$$

The answer is f(0, +-1) = 2e^-1

Again, have no idea how I missed this since this time I don't even have any boundaries to evaluate and the solution is not at infinite.

 

[BvU]

By setting f_x = 0 I get:

You threw away $x=0$ for no reason !

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[BvU]

So my critical points are (1, 0) and (-1, 0)

And then you don't check if they are extrema instead of saddle points !

[edit]just so you know: I'm not trying to be harsh, just try to help .

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[Addez123]

I did forget to check x = 0, but that yields $$y = 1/\sqrt{2}$$ which results in e^-1/2 so it's still not the maximum.

I didn't check for saddlepoints because I am just looking for max and min. If the value is the biggest value it can't be a saddle point eitherway (because I check the limits as x^2+y^2 = r approach infinity aswell).

Also no offence taken :) Feel free to call me stupid aslong as you provide a solution :p

 

[Mark44]

I did forget to check x = 0, but that yields $y = 1/\sqrt{2}$ which results in e^-1/2 so it's still not the maximum.

If x = 0, and $2y^2 = 2$, then both $f_x$ and $f_y$ are zero. Note that $2y^2 = 2$ has two solutions.

 

[BvU]

I did forget to check x = 0, but that yields $$y = 1/\sqrt{2}$$ which results in e^-1/2

Oh, does it ? Can you show me ? I get something else, from
$$f_y = (-2yx^2 + 4y - 4y^3)e^{-x^2 -y^2}\Rightarrow\quad f_y = 0\quad \Leftrightarrow \ \ ...$$

I didn't check for saddlepoints because I am just looking for max and min.

I claim y = 0 yields a minimum (0,0) and two saddlepoints.

If the value is the biggest value it can't be a saddle point either way (because I check the limits as x^2+y^2 = r approach infinity aswell).

criteria for saddle points are ...

Also no offence taken :) Feel free to call me stupid as long as you provide a solution :p

Good. We'll get there.

Where are you in your curriculum ? (I have a reason for asking )

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[Addez123]

When looking at f_x = 0, I deduct that x₁ = 0
Continuing working on f_x I get the expression
$$x^2 = 1 - 2y^2$$
I plug in my x₁ = 0 into this and I get
$$y = 1/\sqrt{2}$$

If I plug in x = 0 into f_y instead, then you get the equation $$4y -4y^3 = 0$$
and from there u get y = +-1

That was what I missed!
It was just not obvious since I plugged in the equation for x^2 into f_y and never thought of testing x₁ aswell!

I'm at maybe 3rd year of university, but I am just studying up a few courses I failed. I don't do school fulltime. What be the reason for asking?

 

[BvU]

What be the reason for asking?

I was wondering if I would do much harm if I showed you a kind of cheat:

On Wolframalpa you can enter the function expression :

(x^2+2y^2)exp(-(x^2+y^2))​

and it gives you a 3d plot and a contour plot of the function plus expressions for x and y derivatives, as well as a list of extrema

((x^2+2y^2)exp(-(x^2+y^2)))' gives you a 3d and a contour plot of the x derivative

d/dy ((x^2+2y^2)exp(-(x^2+y^2))) gives you a 3d and a contour plot of the y derivative

Nifty tool ! Didn't have that when I was in 3rd year, half a century ago

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[Addez123]

Hahah, I've been using that since day one xD
The answers are always in your book though. But yeah, wolfram is useful!