91Ó°ÊÓ

Suppose$\mathit{f}$and$\mathit{g}$are two functions such that they both have continuous partial derivatives.

Let
localid="1664370584970" $\left(x_0,y_0,z_0\right)∈\mathrm{S}:\left\{\left(x,y,z\right):\mathrm{g}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=0\right\}$and$∇G\left(x_0,y_0,z_0\right)≠0$.

If the function $\mathit{f}$contains a local minimum or local maximum at a point ${\mathit{x}}_{\mathbf{0}}\mathbf{,}{\mathit{y}}_{\mathbf{0}}\mathbf{,}{\mathit{z}}_{\mathbf{0}}$then there exists$\mathrm{λ}∈R$such that:

$∇f\left(x_0,y_0,z_0\right)=\mathrm{λ}∇g\left(x_0,y_0,z_0\right)$

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

localid="1664371097882" $\begin{array}{rcl}∇f\left(x,y,z\right)& =& \mathrm{λ}∇g\left(x,y,z\right)\\ g(x,y,z)& =& 0\end{array}$

By solving these equations, we get the values of unknown variables, say $\begin{array}{rcl}& & x,y,z\end{array}$and $\begin{array}{rcl}& & \mathrm{λ}\end{array}$.

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 point $(x,y)$is constrained on the line $2x+3y-4=0$and they need to find the minimize sum of the distances squares $\left(d^2\right)$from the given two points.

So, will proceed in this problem using the Lagrange multiplier method is given below:

$2x+3y-4=0$ …... (1)

$$\begin{gathered} d^2=\left({}_{-}{}^x+\right({}_{-}{}^y+\left({}_{-}{}^x+\right({}_{-}{}^y \\ {d}^2=2(x^2+y^2+1) \end{gathered}$$

Now invoke the Lagrange multiplier method.

$F=2(x^2+y^2+1)+\mathrm{λ}(2x+3y-4)$

Take the partial derivative with respect to $x$and $y$.

$\frac{∂F}{∂x}=4x+2\mathrm{λ}=0$ …... (2)

localid="1664361280294" $\frac{∂F}{∂y}=4y+2\mathrm{λ}=0$ …... (3)

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

With the use of equation (2), $\mathrm{λ}=-2x$.

Substitute for $\mathrm{λ}$in equation (3).

$\begin{array}{rcl}4y+3×(-2x)& =& 0\\ y& =& \frac{3}{2}x\end{array}$

Now, substitute in equation (1).

$\begin{array}{rcl}2×9+3×\left(\frac{3}{2}x\right)-4& =& 0\\ 2x+\frac{9}{2}x& =& 4\\ x& =& \frac{8}{13}\\ y& =& \frac{12}{13}\end{array}$

Thus, the smallest point between the two points is $\begin{array}{rcl}\left(x,y\right)& =& \left(\frac{8}{13},\frac{12}{13}\right)\end{array}$.