We have $f( x_1 ,x_2 ,x_3 ) = x_1^2 +x_2^2 +x_3^2 -6x_1 -8x_2 -10x_3 $

The stationary points are given by $ \frac{∂f}{(∂x_1 )}=0 ,\frac{∂f}{(∂x_2 )}=0,\frac{∂f}{(∂x_3 )}=0$

Now $ \frac{∂f}{(∂x_1 )}=2x_1 - 6 ∴ 2x_1 - 6 = 0 \,\,\ i.e. \,\ x_1 = 3$

$ \frac{∂f}{(∂x_2 )}= 2x_2 – 8 ∴ 2x_2 - 8 = 0 \,\,\ i.e. \,\ x_2 = 4$

$ \frac{∂f}{(∂x_3 )}= 2x_3 – 10 ∴ 2x_3 - 10 = 0 \,\,\ i.e. \,\ x_2 = 5$

Thus X$_0$( 3, 4, 5) is the stationary point.

To check whether the point is a point of maxima or minima ,we consider the Hessian matrix at X$_0$( 3, 4, 5)

H = $ \begin{bmatrix} \frac{∂^2 f}{∂x_1^2}&\frac{∂^2 f}{∂x_1 \ ∂x_2}&\frac{∂^2 f}{∂x_1 \ ∂x_3} \\ \frac{∂^2 f}{∂x_2 \ ∂x_1}&\frac{∂^2 f}{∂x_2^2}&\frac{∂^2 f}{∂x_2 \ ∂x_3} \\ \frac{∂^2 f}{∂x_3 \ ∂x_1}&\frac{∂^2 f}{∂x_3 \ ∂x_2}&\frac{∂^2 f}{∂x_3^2} \end{bmatrix} = \begin{bmatrix} 2&0&0 \\ 0&2&0 \\ 0&0&2 \end{bmatrix} $

The principal minors of this matrix are [2] , $ \begin{bmatrix} 2&0 \\ 0&2 \end{bmatrix}, $ $ \begin{bmatrix} 2&0&0 \\ 0&2&0 \\ 0&0&2 \end{bmatrix} $

The values of their determinants are 2, 4, 8.

Since all the values of determinants are positive, z is minimum at X$_0$( 3, 4, 5)

∴ z$_{min}$ = f( 3, 4, 5) = -50