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Determine the minimum value of the expression $$x^2+y^2+5z^2-xy-3yz-xz+3x-4y+7z$$ where x , y and z are real numbers.

My Solution : Let $f(x,y,z)=x^2+y^2+5z^2-xy-3yz-xz+3x-4y+7z$

I calculated the partial derivative equations as :
$\frac{df}{dx}=2x-y-z+3=0$
$\frac{df}{dy}=2y-x-3z-4=0$
$\frac{df}{dz}=10z-3y-x+7=0$

Solving above equations, I got $x=-4$,$y=-3$,$z=-2$

Then calculating further partial derivatives, I got:

$A=\frac{d^2f}{dx^2}=2$
$B=\frac{d^2f}{dy^2}=2$
$C=\frac{d^2f}{dz^2}=10$
$F=\frac{d^2f}{dydz}=-3$
$G=\frac{d^2f}{dxdz}=-1$
$H=\frac{d^2f}{dxdy}=-1$

Now, matrices :

$$A_{1} = \begin{vmatrix} A & H \\ H & B \end{vmatrix} = 3$$

$$A_{2} = \begin{vmatrix} A & H & G \\ H & B & F \\ G & F & C \end{vmatrix} = 4$$

$A$, $A_{1}$ and $A_{2}$ are positive. So, $f(x,y,z)$ is minimum at $x=-4$,$y=-3$,$z=-2$.

So, $f(x,y,z)$ at $x=-4$,$y=-3$,$z=-2$ => $f(x,y,z)=-7$.

Answer : -7.

Is this solution correct? Please advise.

Math Tise
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    Please use MathJax to make the question more easily readable. In particular, I really don't understand what you're saying in the matrix part. – Matti P. Jul 10 '18 at 05:15
  • @MattiP. ... I tried using MathJax, but unfortunately, it's not working on my PC for some reason. – Math Tise Jul 10 '18 at 05:19
  • Try at least putting dollar signs around the first equation and update the post. See what happens .... – Matti P. Jul 10 '18 at 05:23
  • There will only be one extremum so to show your solution is a minimum evaluate $f$ somewhere else and if it is larger than -7 you have a minimum. Recommend looking at $x=y=z=0$ – user121049 Jul 10 '18 at 07:47
  • @user121049 ...thanks for the suggestion. I am getting all values of f(x,y,z) as more than -7; for x=-4,y=-3,z=-2. Please advise if my solution is accurate. Thanks. – Math Tise Jul 10 '18 at 11:57

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