Classifying Critical Points

1. Mar 8, 2009

MathNoob123

1. The problem statement, all variables and given/known data
f(x,y)=xy+(144/x)+(12/y)
f has a relative minimum at ( , , )

2. Relevant equations

Partial derivatives

3. The attempt at a solution
fx=y-(144/x^2)
fy=x-(12/y^2)

I setted these to zero. Multiplies both sides so there is no fraction.

Came out to be something like~~~ x2y-144=0
xy2-12=0

then did the factoring stuff and got

x2-12xy2=0

This tells me that the critical point is (0,0), but this is the wrong answer.

PLEASE TELL ME WHAT I AM DOING WRONG.

THANK YOU SO MUCH TO THOSE WHO HELP. THIS FORUM IS HELPING ME BECOME A BETTER MATHMATICIAN AND THEREFORE, MY TEST GRADES ARE GOING UP. THANK YOU.

2. Mar 8, 2009

gabbagabbahey

Your partial derivatives and general procedure are correct; but your attempt to solve for x and y is very confusing...try solving the f_y=0 equation for x, and substitute that result into your f_x=0 equation and solve for y....

3. Mar 9, 2009

lanedance

just to add - there can be (and is in this case) more than one stationary point

as you mentioned (0,0) is a stationary point, but that doesn't automatically make it an extremum - need to look at the 2nd order partial dervatievs to determine that
Edit - (0,0) is not a critical point

Last edited: Mar 9, 2009
4. Mar 9, 2009

HallsofIvy

(0, 0) most certainly is NOT a stationary point- it's not even a point on the graph. f is not defined at (0,0).
And I get only one critical point.

5. Mar 9, 2009

lanedance

yeah - looked at that a little too quickly, sorry for the mis-steer