What is the Mistake in Finding the Critical Point for f(x,y)=xy+(144/x)+(12/y)?

[MathNoob123]

Homework Statement

f(x,y)=xy+(144/x)+(12/y)
f has a relative minimum at ( , , )

Homework Equations

Partial derivatives

The Attempt at a Solution

f_x=y-(144/x^2)
f_y=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.

 

[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...

 

[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

 

[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.

 

[lanedance]

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