# Thread: Find the absolute extrema with boundary

1. ## Find the absolute extrema with boundary

We've work on these types of problems in class; however, not with e so im having trouble finding the critical points.

f(x,y)=xye^(y-(x^2)) on the rectangular region bounded by x=0, x=5, y=0, y=-3

we have to find the critical points and we also have to list the corner points and list our min and max

i differentiated it and i think im okay so far

F with respect to x = ye^(y-(x^2))+xy(e^(y-(x^2))(-2x)
F with respect to y = xe^(y-(x^2))+xy(e^(y-(x^2))(1)

2. ## Re: Find the absolute extrema with boundary

Partial derivatives are correct. Set them to zero and solve. Btw, notice the exponential term can't be zero, so the other factor is.

Then differentiate again to decide whether your x and y are a min or max, as explained here:

Pauls Online Notes : Calculus III - Relative Minimums and Maximums

Check here:

f(x,y) = x y e^(y - x^2) - Wolfram|Alpha

3. ## Re: Find the absolute extrema with boundary

There is a theorem saying that any continuous function has both maximum and minimum values on a closed and bounded interval and if they occur in the interior, the partial derivatives are 0 there. However, it is not necessarily true that the maximum and/or minimum will occur on the interior. Even if the partial derivatives are 0, the point may not be a max or min. You need to look on the boundaries also. For example, on the boundary line x= 5, the function is f(5, y)= 5ye^(y- 25).

And, while the "second derivative test" will tell whether or not you have a local max or min, it does not generally tell you if you have a global max or min. Instead, evaluate the function at each possible max or min and compare the values.