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.