In Exercises 39–43, find the point on the given surface closest to the specified point.

$xyz=1\text{and}(0,0,0)$

The given point is the origin and the constraint is$xyz=1.$

It is given that the point is $\left(0,0,0\right)$so the distance of a point from the origin is$\sqrt{x^2+y^2+z^2}.$

Let the function be the square of the above distance $D(x,y,z)=x^2+y^2+z^2$and the constraint as$g(x,y,z)=xyz-1.$

By using the Lagrange's method, the gradient of the above functions are,

$$\begin{gathered} \mathrm{\nabla }D(x,y,z)=2x\mathbf{i}+2y\mathbf{j}+2z\mathbf{k} \\ \mathrm{\nabla }\mathrm{g}(\mathrm{x},\mathrm{y},\mathrm{z})=\mathrm{yz}\mathbf{i}+\mathrm{xz}\mathbf{j}+\mathrm{xy}\mathbf{k} \end{gathered}$$

Now, the system of the equation is

$$\begin{gathered} \mathrm{\nabla }D(x,y,z)=\lambda \mathrm{\nabla }g(x,y,z) \\ 2x\mathbf{i}+2y\mathbf{j}+2z\mathbf{k}=\lambda (yz\mathbf{i}+xz\mathbf{j}+xy\mathbf{k}) \end{gathered}$$

Here,$2x=\lambda yz,2y=\lambda xz\text{and}2z=\lambda xy.$

The values of $\lambda ,$we get are $\frac{2x}{yz}=\frac{2y}{xz}=\frac{2z}{xy}=\lambda .$

So,$x^2=y^2\text{and}{z}^2=y^2.$

Substitute the above values in the given constraint,

$$\begin{gathered} xyz=1 \\ \mathrm{Square}\mathrm{both}\mathrm{sides} \\ {x}^2{y}^2{z}^2=1 \\ {y}^6=1 \\ y=\pm 1 \\ \mathrm{When}\mathrm{y}=-1,\mathrm{x}=\pm 1,\mathrm{z}=\pm 1\mathrm{and}\mathrm{when}\mathrm{y}=1,\mathrm{x}=\pm 1,\mathrm{z}=\pm 1 \end{gathered}$$

Thus, the points where the extreme value may appear are $$\begin{gathered} (-1,-1,-1),(-1,-1,1),(1,-1,-1),(1,-1,1), \\ (-1,1,-1),(-1,1,1),(1,1,-1),(1,1,1). \end{gathered}$$

To find the point closest to the specified point, let's put the points in the function $D(x,y,z)=x^2+y^2+z^2.$

So, the points we get are all the same, thus the value we get is $3.$

Thus, all the points on the surface are closest to the origin.