- #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?

$$f_x = (2x - x^2)e^{-x-y}$$

$$f_y = (1 - y)e^{-x-y}$$

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?