Question

Calculus 3: Absolute Maximum & Minimum

f(x,y)=2+2x+2y-x^2-y^2 with a closed triangular region with vertices at (0,0), (0,6), and (6,0)

I have found the critical point of the interior of the triangle but to find the max and min of the boundaries I just plugin (0,0), (0,6), and (6,0) into f(x,y) to figure it out correct? If so, can the same be said for a rectangle?

Answer 1

Also, is the absolute maximum f(1,1)=4 and absolute global minimum f(0,9) and f(9,0)?

Answer 2

you need to determine your partials right?

Answer 3

f(x,y)=2+2x+2y-x^2-y^2

Fx = 2 - 2x = 0, x = 1
Fxx = -2
Fxy = 0

Fy = 2 - 2y = 0; y = 1
Fyy = -2

D > 0 .... but i tend to forget what that tells us about a possible saddle point :/

Answer 4

If D>0 and Fxx >0 then there is a relative minimum
If D>0 and Fxx < 0 then there is a relative maximum

so i believe that tells us that 1,1 is a relative max, is that within or on our boundary region?

Answer 5

you have to define the line between the vertexes, and evaluate the function along that line ...

Answer 6

(0,0) (0,6)

x=0, y = 0 to 6

f(0,y)=2+2y-y^2

Fy = 2-2y= 0 at y=1 again, so the point (0,3) is a critical point to test ... if im remembering it correctlyt

Answer 7

So you have to check along the lines and vertices correct?

Answer 8

yes

Answer 9

Ohhhhhh! Seems like more work but definitely doable now =)

Answer 10

One moment, I'm trying to trying a PDF that found the maximum and minimum but the professor didn't bother with the lines and only focus on the vertices.