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.
  • #1
Addez123
199
21
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?
 
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  • #2
Addez123 said:
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?##
 
  • #3
First of all, your derivatives are incorrect.

Addez123 said:
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).
 
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  • #4
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:
  • #5
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.
 
  • #6
Addez123 said:
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.
 
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  • #7
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!
 

Related to Can't find the correct max on this surface

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.

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