Question: Find the shortest distance from the origin to each of the following quadric surfaces. Hint: See Example 3 above.

$\mathbf{4}{\mathit{y}}^{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathit{z}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathit{x}\mathbf{=}\mathbf{18}$

The given function is$4y^2+2z^2+3x=18$whose minimum distance from origin is required.

To find the minimum distance of the origin from the point (x , y , z), the distance$\mathbf{d}\mathbf{=}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}$ is to be minimised.

Let $f=d^2=x^2+y^2+z^{2}andalso\varphi \left(x,y,z\right)=4y^2+2z^2+3xy$.

Using the Lagrange multiplier $\mathrm{F}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)={\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2+\mathrm{\lambda }\left(4{\mathrm{y}}^2+2{\mathrm{z}}^2+3\mathrm{xy}\right)$is to be minimised.

Differentiate $\mathrm{F}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)$with respect to z and equate to 0 and solve using .

$$\begin{gathered} \frac{\partial \mathrm{F}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)}{\partial z}=2z+\lambda \left(4z\right) \\ 2z+\lambda \left(4z\right)=0 \\ z\left(2\lambda +1\right)=0 \\ z=0\mathrm{or}\mathrm{\lambda }=\frac{1}{2} \end{gathered}$$

Differentiate $\mathrm{F}(\mathrm{x},\mathrm{y},\mathrm{z})$with respect to y and equate to 0 and solve using $\lambda =-\frac{1}{2}$.

$$\begin{gathered} \frac{\mathrm{\lambda F}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)}{\partial \mathrm{y}}=2\mathrm{y}+\mathrm{\lambda }\left(8\mathrm{y}+3\mathrm{x}\right) \\ 2\mathrm{y}+\mathrm{\lambda }\left(8\mathrm{y}+3\mathrm{x}\right)=0 \\ 2\mathrm{y}-\frac{1}{2}\left(8\mathrm{y}+3\mathrm{x}\right)=0 \\ -2\mathrm{y}-\frac{3}{2}\mathrm{x}=0 \end{gathered}$$

Differentiate F(x,y,z) with respect to x and equate to 0 and solve using $\lambda =-\frac{1}{2}$

$$\begin{gathered} \frac{\partial F\left(x,y,z\right)}{\partial x}=2x+\lambda \left(3y\right) \\ 2x-\frac{1}{2}\left(3y\right)=0 \\ 2x-\frac{3}{2}y=0 \end{gathered}$$

It can be observed that when $\lambda =-\frac{1}{2}$, x = y = 0. Substitute x = y = 0 into to find $4y^2+2z^2+3xy=18$the value of z.

$$\begin{gathered} 4{\left(0\right)}^2+2{\mathrm{z}}^2+3\left(0\right)\left(0\right)=18 \\ {\mathrm{z}}^2=9 \\ \mathrm{z}=\pm 3 \end{gathered}$$

When z = 0,Find the value of $\lambda$.

$$\begin{gathered} \lambda =-\frac{2y}{8y+3x}=-\frac{2x}{3y} \\ \frac{2y}{8y+3x}=-\frac{2x}{3y} \\ 3x^2+8xy-3y^2=0 \\ x=-3y,\frac{y}{3} \end{gathered}$$

Substitute $x=-3y,\frac{y}{3}$and z = 0 into $4y^2+2z^2+3xy$to find the value of y.

$$\begin{gathered} 4y^2+2{\left(0\right)}^2+3\left(-3y\right)y=18 \\ 4y^2-9y^2=18 \\ -5y^2=18 \end{gathered}$$

Thus x = -3y is not feasible.

$$\begin{gathered} 4y^2+2{\left(0\right)}^2+3\left(\frac{y}{3}\right)=18 \\ 4y^2+y^2=18 \\ 5y^2=18 \\ y=\pm \sqrt{\frac{18}{5}} \end{gathered}$$

Thus, the points obtained are $\left(0,0,\pm 3\right),\left(-\sqrt{\frac{2}{5}},\sqrt{\frac{18}{5}},0\right)\mathrm{and}\left(\sqrt{\frac{2}{5}},-\sqrt{\frac{18}{5}},0\right)$.

Find the distance of the obtained points from the origin, that is $\mathrm{d}=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2}$.

$$\begin{gathered} \mathrm{d}=\sqrt{{\left(0\right)}^2+{\left(0\right)}^2+{\left(\pm 3\right)}^2} \\ =3 \\ \mathrm{d}=\sqrt{{\left(-\sqrt{\frac{2}{5}}\right)}^2+{\left(-\sqrt{\frac{18}{5}}\right)}^2+{\left(0\right)}^2} \\ =2 \\ \mathrm{d}=\sqrt{{\left(-\sqrt{\frac{2}{5}}\right)}^2+{\left(-\sqrt{\frac{18}{5}}\right)}^2+{\left(0\right)}^2} \\ =2 \end{gathered}$$