Find the point on $\mathbf{2}\mathit{x}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{+}\mathit{Z}\mathbf{-}\mathbf{11}\mathbf{=}\mathbf{0}$for which $\mathbf{4}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{+}{\mathit{z}}^{\mathbf{2}}$is a minimum.

Suppose f and g are two functions such that they both have continuous partial derivatives.

Let $\mathbf{(}{\mathit{x}}_{\mathbf{0}}\mathbf{,}{\mathit{y}}_{\mathbf{0}}\mathbf{,}{\mathit{z}}_{\mathbf{0}}\mathbf{)}\mathbf{\in }\mathit{S}\mathbf{:}\mathbf{=}\mathbf{\left\{}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{:}\mathbf{g}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{\right\}}\mathit{a}$and$\mathbf{\nabla }\mathit{g}\mathbf{(}{\mathit{x}}_{\mathbf{0}}\mathbf{,}{\mathit{y}}_{\mathbf{0}}\mathbf{,}{\mathit{z}}_{\mathbf{0}}\mathbf{)}\mathbf{\ne }\mathbf{0}$.

If the function fcontains a local minimum or local maximum at a point $\mathbf{(}{\mathit{x}}_{\mathbf{0}}\mathbf{,}{\mathit{y}}_{\mathbf{0}}\mathbf{,}{\mathit{z}}_{\mathbf{0}}\mathbf{)}$then there exists $\mathit{\lambda }\mathbf{\in }\mathit{R}$such that:

$\mathbf{\nabla }\mathit{f}\mathbf{(}{\mathit{x}}_{\mathbf{0}}\mathbf{,}{\mathit{y}}_{\mathbf{0}}\mathbf{,}{\mathit{z}}_{\mathbf{0}}\mathbf{)}\mathbf{=}\mathit{\lambda }\mathbf{\nabla }\mathit{g}\mathbf{(}{\mathit{x}}_{\mathbf{0}}\mathbf{,}{\mathit{y}}_{\mathbf{0}}\mathbf{,}{\mathit{z}}_{\mathbf{0}}\mathbf{)}$

To determine the extremism points, we generally consider the below equations:

$\begin{array}{l}\mathbf{\nabla }\mathbf{f}\left(x,y,z\right)\mathbf{=}\mathbf{\lambda }{\mathbf{\nabla }}_{\mathbf{g}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{]}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{g}\left(x,y,z\right)\mathbf{=}\mathbf{0}\end{array}$

By solving these equations, we get the values of unknown variables, saylocalid="1659163338102" $\mathit{x}\mathbf{,}\mathbf{\hspace{0.33em}}\mathit{y}\mathbf{,}\mathbf{\hspace{0.33em}}\mathit{z}\mathbf{}$and $\mathit{\lambda }$.

Thus, we will get the local extremism points through the solutions of the above set of equations.

From the given information, it is a typical optimization problem with constraint, where the points $(x,y,z)$are constraint to move on the plane $2x+3y+z-11=0.$

So, will use Lagrange multiplier method given below:

$2x+3y+z-11=0.$ …... (1)

Now invoke the Lagrange multiplier method.

$F=4x^2+y^2+z^2+\lambda (2x+3y+z-11)$

Take the partial derivative with respect tox, y, andz .

$\frac{\partial F}{\partial x}=8x+2\lambda =0$ …... (2)

$\frac{\partial F}{\partial y}=2y+3\lambda =0$ …... (3)

$\frac{\partial F}{\partial z}=2z+\lambda =0$ …... (4)

$\frac{\partial F}{\partial l}=r(r+2\lambda )=0$

Now with three partial derivative equations and the original constraints equation will be solve for unknown proportions.

By the use of equation (4), $\lambda =-2z$.

Substitute for $\lambda$in equation (3).

$$\begin{gathered} 2y+3x-2z=0 \\ y=3z \end{gathered}$$

Now, substitute in equation (2).

$$\begin{gathered} 8x+2\times -2z=0x \\ x=\frac{z}{2} \end{gathered}$$

Finally, substituting in equation (1) for x and y.

$$\begin{gathered} 2\times \left(\frac{z}{2}\right)+3\times \left(3z\right)+z-11=0 \\ z=1 \\ y=3 \\ x=\frac{1}{2} \end{gathered}$$

Thus, points of the equations are $x=\frac{1}{2},y=3,z=1$.