Question

Find the absolute max and min of f on the set D. f(x,y)=x+y-xy, D is the closed triangular region with vertices (0,0), (0,2), (4,0)

I know that a critical point is: f(1,1)=1 and f(0,y)=y, f(x,0)=x, and f(x,(-1/2)x+2)=((x^2)/2)+(x/2). But I have no idea what to do from there. I just have no idea how to find the max and min over the closed region.

Answer 1

@phi @satellite73

Answer 2

Paul's notes goes over problems like these
http://tutorial.math.lamar.edu/Classes/CalcIII/AbsoluteExtrema.aspx

Answer 3

Ok, so \[f_x=1-y...y=1 \\ f_y=1-x... x=1\] If I plug y=1 into f_x I get 1-1=0, so would the critical point actually be (0,1)?

And plugging x=1 from the second equation into f_y I get 1-1=0, so would that critical point be (1,0)?

Answer 4

When I replace y with -x/2 + 2 in your f(x,y) I get
(1/2) x^2 - (3/2) x +2
this has a critical point at x-3/2 =0 or x= 3/2

Answer 5

The critical points require both fx and fy be 0 (otherwise one of the dimensions is still peaking (or going down))

Answer 6

Awesome! So, these would be the critical points. Now how would I test them to check if they are a max or min? I know for unbounded sets I would use f_xx+f_yy-(f_xy)^2.

Would I use that here?

Answer 7

I would reason like this. If the critical point (1,1) within the region is not a max then the max must occur on a boundary. (in other words if we don't see a peak within the region)
check all the boundaries. on x=0 f(x,y)=y which is a max at y=2
on y=0 , f(x,y)= x and it has a max at x=4

Answer 8

along the "hypotenuse" it max or mins at x=3/2

I would evaluate to find the value of the function at
(1,1)
(3/2, 7/8)
(0,0)
(0,2)
(4,0)

Answer 9

How do I check if it's a max or min? I was told that the typical formula...
f_xx+f_yy-(f_xy)^2
Doesn't work for these?

Answer 10

*(3/2, 7/8) should be (3/2, 5/4)

Answer 11

I think that formula tells you if you have a min or max based on what the 2nd derivatives are doing. But on the boundary the function could still be going up, so it is not really a max. It is a max in the region, but goes higher outside the region.

So the idea is to just evaluate the function at a few "interesting" points

Answer 12

Would I just plug those values into the main function to see which gives the largest and smallest values?

Answer 13

yes. It makes sense.

(1,1) will be a max, min or an inflection. Just find f(1,1)
(0,0) will be a max or min (because the function has no critical points along the line x=0)
(0,2) ibid
(4,0) ibid
(3/2, 5/4) is a critical point along the hypotenuse.

Answer 14

Ok, so...
f(x,y)=x+y-xy
f(1,1)=1+1-1=1
f(0,0)=0
f(0,2)=2
f(4,0)=4
f(3/2,5/4)=7/8

The largest is f(4,0), so that would be the max and f(0,0) would be the min. How would I check for a saddle point?

Answer 15

according to http://tutorial.math.lamar.edu/Classes/CalcIII/RelativeExtrema.aspx
(see "fact" just before example 1)
use your Dxx + Dyy - (Dxy)^2
on your only critical point (1,1)
if <0 it is a saddle point
I think it passes this test, so (1,1) is a saddle

Answer 16

Thank you!

Lastly, how did you get 5/4 for f(3/2,5/4)? When I plugged 3/2 back into
(1/2) x^2 - (3/2) x +2 I got 7/8 like you did the first time?

Answer 17

because we are looking for "interesting" points along the line
y= (-1/2) x + 2 (the boundary line)
once we find x and y, we can evaluate the function f(x,y) to find its value at that point on the boundary.

Answer 18

btw, (3/2, 5/4) looks like a local min of f(x,y) as we move from (0,2) to (4,0) along that line

Answer 19

Ok, I understand how you got 5/4 now. Did you find out it was local min from the Dxx + Dyy - (Dxy)^2 equation?

Answer 20

no, I looked at the values of f(0,2) and f(4,0)

Answer 21

Thank you!