Find the shortest distance from the origin to the surface $\mathit{x}\mathbf{=}\mathit{y}\mathit{z}\mathbf{+}\mathbf{10}$.

The given equation is $x=yz+10$.

Let $\left(x,y,z\right)$be any point on the surface.

The distance between the origin$\left(0,0,0\right)$ to the point $\left(x,y,z\right)$is:

$$\begin{gathered} d^2={\left(x-0\right)}^2+{\left(y-0\right)}^2+{\left(z-0\right)}^2 \\ {d}^2=x^2+y^2+z^2 \end{gathered}$$

Consider the function $f\left(x,y,z\right)=x^2+y^2+z^2...\left(1\right)$.

Now, use the Lagrange multipliers as:

Differentiate (1) with respect to $x$, then to get$f_x\left(x,y,z\right)=2x$ .

Differentiate (1) with respect to $y$, then to get$f_y\left(x,y,z\right)=2y$ .

Differentiate (1) with respect to $z$, then to get $f_z\left(x,y,z\right)=2z$.

Differentiate $g\left(x,y,z\right)=x-yz-10=0$with respect to$x$ , then to get$g_x\left(x,y,z\right)=1$ .

Differentiate$g\left(x,y,z\right)=x-yz-10=0$ with respect to $y$, then to get $g_y\left(x,y,z\right)=-z$.

Differentiate $g\left(x,y,z\right)=x-yz-10=0$with respect to$z$ , then to get$g_z\left(x,y,z\right)=-y$ .

Solve the equation as follows:

$$\begin{gathered} f_x=\lambda g_x \\ 2x=\lambda \times 1 \\ x=\frac{\lambda }{2}...\left(2\right) \end{gathered}$$

Now, solving as:

$$\begin{gathered} f_y=\lambda g_y \\ 2y=\lambda \times \left(-z\right) \\ \frac{y}{z}=-\frac{\lambda }{2}...\left(3\right) \end{gathered}$$

Solving further as:

$$\begin{gathered} f_z=\lambda g_z \\ 2z=\lambda \times \left(-y\right) \\ \frac{y}{z}=-\frac{2}{\lambda }...\left(4\right) \end{gathered}$$

And $x-yz-10=0...\left(5\right)$.

From (3) and (4) to get the$\lambda$value is:

$$\begin{gathered} -\frac{\lambda }{2}=-\frac{2}{\lambda } \\ {\lambda }^2=4 \\ \lambda =\pm 2 \end{gathered}$$

Then the value of $x$is:

$$\begin{gathered} x=\frac{\lambda }{2} \\ x=\frac{\pm 2}{2} \\ x=\pm 1 \end{gathered}$$

By observing (3) and (4) equations, write:

$$\begin{gathered} \frac{y}{z}=-\frac{\lambda }{2} \\ \lambda =-\frac{2y}{z} \end{gathered}$$

And:

$$\begin{gathered} \frac{y}{z}=-\frac{2}{\lambda } \\ \lambda =-\frac{2z}{y} \end{gathered}$$

From these two functions, obtain:

$$\begin{gathered} -\frac{2y}{z}=-\frac{2z}{y} \\ {y}^2=z^2 \end{gathered}$$

Now, use the constraint $x-yz-10=0$as:

$$\begin{gathered} yz=x-10 \\ {y}^2{z}^2={\left(x-10\right)}^2 \end{gathered}$$

When $x=1$as:

$$\begin{gathered} y^2{z}^2={\left(x-10\right)}^2 \\ {y}^2{z}^2=81 \end{gathered}$$

When$x=-1$as:

$$\begin{gathered} y^2{z}^2={\left(x-10\right)}^2 \\ {y}^2{z}^2=121 \end{gathered}$$

When $y^2=z^2$, then simplify as follows:

$$\begin{gathered} y^2{z}^2=81 \\ {y}^4=121 \\ y=\pm \sqrt{11} \\ z=\pm \sqrt{11} \end{gathered}$$

$$\begin{gathered} y^2{z}^2=81 \\ {y}^4=121 \\ y=\pm \sqrt{11} \\ z=\pm \sqrt{11} \end{gathered}$$

The corresponding points are:

$\left(-1,\sqrt{11},\sqrt{11}\right),\left(-1,\sqrt{11},-\sqrt{11}\right),\left(-1,-\sqrt{11},-\sqrt{11}\right),\left(-1,-\sqrt{11},\sqrt{11}\right)$

.

So, to find the extreme value of the function$f\left(x,y,z\right)$ , substitute the points in equation (1).

When :$\left(x,y,z\right)=\left(1,3,3\right)$

$$\begin{gathered} f\left(x,y,z\right)=x^2+y^2+z^2 \\ f\left(x,y,z\right)={\left(1\right)}^2+{\left(3\right)}^2+{\left(3\right)}^2 \\ f\left(x,y,z\right)=1+9+9 \\ f\left(x,y,z\right)=19 \end{gathered}$$

That is:

$$\begin{gathered} d^2=f\left(x,y,z\right) \\ {d}^2=19 \\ d=\pm \sqrt{19} \\ d=\pm 4.36 \end{gathered}$$

When :$\left(x,y,z\right)=\left(-1,\sqrt{11},\sqrt{11}\right)$

$$\begin{gathered} f\left(x,y,z\right)=x^2+y^2+z^2 \\ f\left(x,y,z\right)={\left(-1\right)}^2+{\left(\sqrt{11}\right)}^2+{\left(\sqrt{11}\right)}^2 \\ f\left(x,y,z\right)=1+11+11 \\ f\left(x,y,z\right)=23 \end{gathered}$$

That is:

$$\begin{gathered} d^2=f\left(x,y,z\right) \\ {d}^2=23 \\ d=\pm \sqrt{23} \\ d=\pm 4.796 \end{gathered}$$

Similarly for as:$\left(x,y,z\right)=\left(-1,\sqrt{11},-\sqrt{11}\right),\left(-1,-\sqrt{11},-\sqrt{11}\right),\left(-1,-\sqrt{11},\sqrt{11}\right)$

$$\begin{gathered} d=\pm \sqrt{23} \\ d=\pm 4.796 \end{gathered}$$

Take the only positive value because the distance never negative.

Therefore, the minimum distance is$d=\sqrt{19}=4.36$ .