# Can't find the correct max on this surface

In summary, the author re-did their derivatives and found a new critical point at (1/2, 3/4) when f_x, f_y = 0.
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?

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?

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).

Orodruin said:
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)

Last edited:
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.

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.

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!

## 1. What does "finding the correct max on this surface" mean?

Finding the correct max on a surface refers to determining the highest point or value on a given surface or function.

## 2. Why is it important to find the correct max on a surface?

Finding the correct max on a surface is important in various fields of science, such as physics and engineering, as it allows for accurate predictions and calculations.

## 3. What factors can affect the accuracy of finding the correct max on a surface?

The accuracy of finding the correct max on a surface can be affected by the resolution of the data, the complexity of the surface, and the chosen method of analysis.

## 4. How can I improve my ability to find the correct max on a surface?

Improving your ability to find the correct max on a surface can be achieved through practice and understanding of mathematical concepts, as well as utilizing advanced computational tools.

## 5. Are there any common mistakes made when trying to find the correct max on a surface?

Yes, some common mistakes include using incorrect units, not considering all variables, and not taking into account the limitations of the chosen method of analysis.

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