91Ó°ÊÓ

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,

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

Now, the system of the equation is

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

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

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

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

Substitute the above values in the given constraint,

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

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.