Question

Find absolute min/max values of f(x,y) = 4 + 2x^2 + y^2 On Given set R = {(x, y): -1<=x<=1, -1<=y<=1 }

Answer 1

find the critical point of f(x,y) that is in the region, and then find the criticle points on the boundary

Answer 2

Ok Sourwing lemme show you what I have.....

Answer 3

Your equation describes an elliptic paraboloid. Its absolute minimum is at the origin, i.e. at (0,0), and has the value f(0,0) = 4 there.

The absolute maximum is at FOUR places, the corners of your rectangular region, i.e. at (1,1), (1,-1), (-1,1), and (-1,-1). The value is f(1,1) = 4 + 2(1)^2 + (1)^2 = 7.

Try graphing the function.

Answer 4

Find CP of f(x,y):

fx = 4x fy = 2y

set each = 0
4x = 0 x = 0 4y=0 y=o

CP at (0,0)

Answer 5

good f(0,0) = 4

Answer 6

I am not sure how to find critical points of R though...

Answer 7

suppose x = 1, what is f(1,y) ?

Answer 8

f(1, y) = 4 + 2(1)^2 + y^2