Find the critical point of f(x,y)=x+y+7/xy. I know you find the critical point by setting the gradient equal to the zero vector and solving the system of equations for x and y.

Question

tiffany_rhodes · Answer

are the partial derivatives df/dx = 1 -7y/(xy)^2 and df/dy = 1 -7x/(xy)^2

tiffany_rhodes · Answer

?

johnweldon1993 · Answer

$$\large x + y + \frac{7}{xy}$$ right?

tiffany_rhodes · Answer

yes

johnweldon1993 · Answer

Went through the whole thing to say no...but realized you just wrote it different....but yes that is correct

tiffany_rhodes · Answer

Okay, so you set each of the partial derivatives equal to 0 and then solve for x and y correct?

johnweldon1993 · Answer

Correct

tiffany_rhodes · Answer

to find the critical point(s) of the function?

tiffany_rhodes · Answer

Okay, so I'm having trouble with that. I was just curious about the partial derivatives to make sure I had the equations correct.

phi · Answer

1 -7y/(xy)^2 = 0
means
1 = 7/(x^2 y)
x^2 y = 7

do the same for the other ,  df/dy

johnweldon1993 · Answer

Ah...well as I wrote them out I have them as

$$\large f_x = 1 - \frac{7}{x^2y}$$
ad
$$\large f_y = 1 - \frac{7}{xy^2}$$

So lets do that 'x' out
$$\large 1 = \frac{7}{x^2y}$$
ahh phii beat me to it lol

phi · Answer

you should get
x^2 y = 7
x y^2 = 7
divide these equations to get
x/y= 7
x= 7y    

use that in the second equation
7 y * y^2 = 7
solve for y

tiffany_rhodes · Answer

Wow, I didn't realize the top y cancelled out. That makes things a lot simpler. I guess I've been out of algebra too long. Okay, so I got y = 7/x^2 when I solved the first equation

phi · Answer

yes.  see the post above yours for how to continue

tiffany_rhodes · Answer

Thanks for the help!

tiffany_rhodes · Answer

both of you*