Max/min of multivariate function

1. Nov 18, 2014

Panphobia

1. The problem statement, all variables and given/known data
max/min of
f(x,y) = x + y
constraint xy = 16
3. 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?

2. Nov 18, 2014

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.

3. Nov 18, 2014

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.

4. Nov 19, 2014

Ray Vickson

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.