Consider the function fa y 2y2 2r and let D be the closed r

Consider the function f(a, y) 2+y2 - 2r, and let D be the closed region bounded by the triangle with vertices (2.0), (0.2), and (0,-2). Take it for free that the only critical point for f in D is (1,0) and that f(1,0) =-1. Find the absolute minimum and the absolute maximum value for f on D. Hint for the analysis of the boundary: fla.2-x) = f(x,x-2). T, x2 SHOW YOUR WORK

Solution

z = x^2 + y^2- 2x
zx = 2x - 2 = 0 ----> x = 1

zy = 2y = 0
y = 0

So, only critical is (1,0)
which does lie within that triangle

Boundary :
(2,0) to (0,2) line is given by x + y = 2
So, y = 2 - x

So, z = x^2 + (2-x)^2 - 2x
z = x^2 + 4 + x^2 - 4x - 2x
z = 2x^2 - 6x + 4
zx = 4x- 6 = 0
x = 3/2

So, when x = 1.5, we get y = 2-1.5 --> y = 0.5

So another critical is (1.5 , 0.5)

Now, (2,0) and (0,-2) is the line
x - y = 2
y = x -2
z = x^2 + y^2- 2x
z = x^2 + (x - 2)^2 - 2x
z = 2x^2 - 6x + 4
Again we get (1.5 ,0.5)

So, the criticals are
(1,0)
(1.5,0.5)
(2,0)
(0,2)
(0,-2)

Now, we find the fn value at all 5 points :
f = -1 when 1,0
f = -0.5 when 1,5.0.5
f = 0 when 2,0
f = 4 when 0,2
f = 4 when 0,-2

So, abs max = 4 at (0,2) or (0,-2)
abs min = -0.5 at (1.5 , 0.5)