Question

F(x,y) = x^2 + y^2 - 2x, D is the closed triangular region with vertices (0,0), (0,2), and (0,-2). Find the absolute maximum and minimum values of f on the set D. I know the maximum and minimum are f(0,+-2) = 4, minimum f(1,0) = -1. I need a step by step process on how to get to this answer

Answer 1

The absolute maximum value is f(0,±2) = 4 and the absolute minimum value is f(1,0) = -1.

The given function is  F(x,y) = $x^{2}$ + $y^{2}$ - $2x$   → 1

D is the closed triangular region with the vertices (0,0), (0,2) and (0,-2).

Differentiate 1 partially with respect to x:

df/dx =  $x^{2}$ + $y^{2}$ - $2x$

$f^{'}$(x) = 2x - 2

Differentiate 1 partially with respect to y:

df/dy = $x^{2}$ + $y^{2}$ - $2x$

$f^{'}$(y)  = 2y

Consider $f^{'}$(x) = 0 and $f^{'}$(y) = 0 to find the critical points.

$f^{'}$(x) ⇒ 2x -2 =0

x = 1

$f^{'}$(y) ⇒ 2y

y = 0

Thus, the critical point is (x,y) = (1,0).

Calculate the values of f(x,y) at (0,0), (0,2), (0,-2) and (1,0).

At (x,y) = (0,0)

f(0,0) ⇒ 0 + 0 - 0 = 0

At (x,y) = (0,2)

f(0,2) ⇒ 0 + 4 - 0 = 4

At (x,y) = (0,-2)

f(0,-2) ⇒ 0 + 4 - 0 = 4

At (x,y) = (1,0)

f(1,0) ⇒ 1 + 0 - 2 = -1

Therefore, the absolute maximum is f(0,±2) = 4 and the absolute minimum is f(1,0) = -1.

To know more about maximum and minimum problems refer to:

https://brainly.com/question/29409891

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