Is Completing the Square Necessary for Finding Optimal Points in Calculus?

[mlb123]

So I have a function f(x,y,z)=3+4(x^2)-9(z^2)-4xy-12yz. I need to find if (0,0,0) is an optimal number. I know I have to get it into the form of 3+(___)^2+(___)^2+(___)^2 or 3-(___)^2-(___)^2-(___)^2 to find if it's a max or min. But I've been working on this all day to no avail. Any help would be useful and appreciated.

 

[HallsofIvy]

This was posted under "Calculus" so it seems strange that you would be required to use "completing the square" rather than setting the partial derivatives to 0. It that really so?

Assuming that you really need to complete the square, I would write
f(x,y,z)= 3+ 4x^2- 9z^2- 4xy- 12yz= 3+ 4(x^2- yx)- 9(z^2+ (4/3)yz)
f(x,y,z)= 3+ 4(x^2- yx+ y^2/4- y^2/4)- 9(z^2+ (4/3)yz+ (4/9)y^2- (4/9)y^2)
f(x, y, z)= 4(z- y/2)^2- 9(z+ 2y/3)^2+ 3- y^2+ 4y^2
f(x, y, z)= 4(z- y/2)^2- 9(z+ 2y/3)^2+ 3y^2+ 3.

(You are aware, are you not, that there are some critical points that are neither a max nor a min?)

 

[mlb123]

This was posted under "Calculus" so it seems strange that you would be required to use "completing the square" rather than setting the partial derivatives to 0. It that really so?

Assuming that you really need to complete the square, I would write
f(x,y,z)= 3+ 4x^2- 9z^2- 4xy- 12yz= 3+ 4(x^2- yx)- 9(z^2+ (4/3)yz)
f(x,y,z)= 3+ 4(x^2- yx+ y^2/4- y^2/4)- 9(z^2+ (4/3)yz+ (4/9)y^2- (4/9)y^2)
f(x, y, z)= 4(z- y/2)^2- 9(z+ 2y/3)^2+ 3- y^2+ 4y^2
f(x, y, z)= 4(z- y/2)^2- 9(z+ 2y/3)^2+ 3y^2+ 3.

(You are aware, are you not, that there are some critical points that are neither a max nor a min?)

Yes, it is really so. That's why I wasn't understanding the question as much - it didn't want me to answer the question in the manner that I would normally turn to. Thank you for your help, as it confirms the answer that I got a few days ago a little after I posted this.