Answer to Question #107907 in Calculus for annie

Question

Suppose f(x,y)=x2+y2−6x and D is the closed triangular region with vertices (6,0), (0,6), and (0,−6). Answer the following. Find the absolute maximum of f(x,y) on the region D. Find the absolute minimum of f(x,y) on the region D.

Answer 1

f(x,y)=x²+y²−6x

A (6,0)

B (0,6)

C (0, -6)

f_x= 2x-6, f_x= 0 -> x = 3

f_y= 2y, f_y= 0 -> y = 0

Value in critical point f(3,0) = -9

AB: y = 6-x, 0<=x<=6

BC: y = x-6, 0<=x<=6

AC: x= 0, -6<=x<=6

f(x,6-x) = x²+(6-x)²-6x = 2x²-18x+36

f'(x,6-x) = 4x-18, f'(x,6-x) = 0, -> x = 9/2, y = 3/2

Calculate values at the two end points of AB:

f(0,6) = 36

f(9/2,3/2) = -9/2

f(6,0) = 0

Calculate for vector BC:

f(x,x-6) = x²+(x-6)²-6x = 2x²-18x+36

f'(x,6-x) = 4x-18, f'(x,x-6) = 0, -> x = 9/2, y = 3/2

Calculate values at the two end points of BC:

f(0,-6) = 36

f(9/2,3/2) = -9/2

f(6,0) = 0

Finally solve for the AC:

f(0,y)= y²

So f(0,0) = 0

Absolute minimum: f(3,0) = -9

Absolute maximum: f(0,6)=f(0,-6) = 36