Can't find the correct max on this surface

[Addez123]

Homework Statement

Find max and min on
$$f(x,y) = (x^2 + y)e^{-x-y}$$
x >= 0, y >= 0

Relevant Equations

Derivatives and limits

I calculate f_x to be
$$f_x = (2x - x^2)e^{-x-y}$$
$$f_y = (1 - y)e^{-x-y}$$

f_x = 0 when x_1 = 0, x_2 = 2
f_y = 0 when y = 1

This gives two critical points: (0, 1) and (2, 1) which yields e^-1 and 5e^-3 respectively.
I then check the x line and y lines by doing
$$f(x, 0) = x^2e^{-x-y}$$
$$f_x = (2x - x^2)e^{-x-y}$$
Which yields the same critical points as before
$$f(0, y) = ye^{-x-y}$$
$$f_y = (1 - y)e^{-x-y}$$
This also gives same critical points as before.

Basically all my critical points are (0,0), (0, 1), (2,1) which at best gives e^-1 ~ .37
But there's a point f(2, 0) = 4e^-2 ~= .54
How did I not find this point when I checked the y-axis boundary?

 

[fresh_42]

Homework Statement:: Find max and min on
$$f(x,y) = (x^2 + y)e^{-x-y}$$
x >= 0, y >= 0
Relevant Equations:: Derivatives and limits

How did I not find this point when I checked the y-axis boundary?

What about $f_x=0?$

 

[Orodruin]

First of all, your derivatives are incorrect.

How did I not find this point when I checked the y-axis boundary?

Your method is wrong. $y = 1$ is not on the $y$-axis. You do not need the y-derivative to vanish when you are looking for extrema on the y-axis, just the derivative along the boundary (in the case of the y-axis, the derivative in the x-direction).

 

[Addez123]

First of all, your derivatives are incorrect.

I redid them and updated them to:
$$f_x = (2x - x^2 - y)e^{-x-y}$$
$$f_y = (1 - y - x^2)e^{-x-y}$$
This gives me a new critical point at (1/2, 3/4), when f_x, f_y = 0

This really doesn't help though.
$$f(x, 0) = x^2e^{-x}$$
$$f_x = (2x - x^2)e^{-x}$$
Going along the lines you still only get (0, 1) and (2,1) points.

I still can't find (2, 0)

 

[fresh_42]

If you look at the boundaries, then you have $x=0$ and $y=0$. Each gives you a function in one variable that you can calculate maximum points of. The minimum is obviously zero, achieved at $(0,0)$ and at infinity. Remains to check whether there could be another maximum inside the surface.

 

[Orodruin]

Going along the lines you still only get (0, 1) and (2,1) points.

First, your $f(x,0)$ cannot depend on y as it does in your post. Second, there is no way you can find (2,1) when considering $f(x,0)$. You can only find points on the $x$-axis and the only relevant derivative for this function is wrt x.

 

[Addez123]

Ahh that was the problem!
I was considering the case when y= 0 yet I thought I was suppose to calculate the y value using f_y.
Now I get the point (2, 0) which gives correct max!

Thanks!