Max/min of multivariate function

[Panphobia]

Homework Statement

max/min of
f(x,y) = x + y
constraint xy = 16

The Attempt at a Solution

With lagrange multipliers I did
$\nabla f = (1,1)$
$\nabla g = (y,x)$
$\nabla f = \lambda \nabla g$
$1 = \lambda y$
$1 = \lambda x$
Since y=0, x=0 aren't a part of xy = 16 I can isolate for lambda

$y = x$
$y^2 = 16$
$y = \pm 4$
$y = 4, x = 4$
$y = -4, x = -4$
$f(4,4) = 8$
$f(-4,-4) = -8$
I got these values, but my answer key says that there are no minimums or maximums, can anyone explain why?

 

[LCKurtz]

The values you have found are relative min/max points as you move along xy=16. But neither are absolute extrema because x+y gets larger and smaller than both then you let either x or y get large positive or negative.

 

[Panphobia]

Yea I understand since as x approaches infinity, y approaches 0, or x approach negative infintiy y approaches 0, so f(x,y) never has a max or min.

 

[Ray Vickson]

Yea I understand since as x approaches infinity, y approaches 0, or x approach negative infintiy y approaches 0, so f(x,y) never has a max or min.

In the positive quadrant $x, y \geq 0$ your constrained $f$ does have a minimum, but no maximum. In the third quadrant $x \leq 0, y \leq 0$ the constrained function has a maximum, but no minimum. If we throw out the information about quadrants then, of course, it is true that the constrained $f$ had neither a maximum nor a minimum.