Question

Do Example 3b using Lagrange multipliers.

Short answer

Hence, the points on the given surface nearest to the origin are $\left(0,0,1\right)\text{and}\left(0,0,-1\right)$.

Step-by-step solution

Lagrange's Multipliers

Lagrange's multipliers is a method of optimization to find the local maxima and minima of the function $f\left(x,y,z\right)$ with constraint $\phi$, which is given by:

$F\left(x,y,z\right)=f+\lambda \phi$

Find derivatives

The given functions are:

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

Now, using Lagrange's multipliers method, we get:

$$\begin{gathered} F\left(x,y,z\right)=f+\lambda \phi \\ =x^2+y^2+z^2+\lambda \left(z^2-x^2\right) \end{gathered}$$

To find the minimum or maximum angle, we have:

$$\begin{gathered} \frac{\partial F}{\partial x}=2x-2x\lambda =0 \\ \frac{\partial F}{\partial y}=2y=0 \\ \frac{\partial F}{\partial z}=2z+2z\lambda =0 \end{gathered}$$

Find the values of variables

From $\frac{\partial F}{\partial x}=0$, we get:

$$\begin{gathered} x=0\text{or}\lambda =1 \\ \Rightarrow z=\pm 1 \\ \text{or}z=0,x=\pm i \end{gathered}$$

From $\frac{\partial F}{\partial y}=0$, we get:

$y=0$

From $\frac{\partial F}{\partial z}=0$, we get:

$$\begin{gathered} z=0\mathrm{or}\lambda =-1 \\ \Rightarrow x=\pm i \\ \mathrm{or}x=0,z=\pm 1 \end{gathered}$$

Taking the real values of points, the point of interest obtained is role="math" localid="1664881443001" $\left(0,0,\pm 1\right)$.

Hence, the points on the given surface nearest to the origin are $\left(0,0,1\right)\mathrm{and}\left(0,0,-1\right)$.