example of using Lagrange multipliersExampleOfUsingLagrangeMultipliers2009-02-06 13:29:142009-02-08 12:39:34ExampleLagrange multiplier methodLagrange multiplier% this is the default PlanetMath preamble. as your knowledge
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One \PMlinkescapetext{way} to determine the perpendicular distance of the parallel planes
$$Ax+By+Cz+D \;=\; 0 \quad \mbox{and} \quad Ax+By+Cz+E \;=\; 0$$
is to use the Lagrange multiplier method.\, In this case we may to minimise the Euclidean distance of a point
\,$(x,\,y,\,z)$\, of the former plane to a (fixed) point \,$(x_0,\,y_0,\,z_0)$\, of the latter plane.\\
Thus we have the equation \,$Ax_0+By_0+Cz_0+E \,=\, 0$\, which we can subtract from the first plane equation, getting
\begin{align}
g \;:=\; A(x-x_0)+B(y-y_0)+C(z-x_0)+D-E \;=\; 0.
\end{align}
This is the (only) constraint equation for minimising the \PMlinkname{square}{SquareOfANumber}
\begin{align}
f \;:=\; (x-x_0)^2+(y-y_0)^2+(z-x_0)^2
\end{align}
of the distance of the points.
The polynomial functions $f$ and $g$ satisfy the differentiability requirements.\, Accordingly, we can find the minimising point\, $(x,\,y,\,z)$\, by considering the system of equations formed by (1) and
\begin{align}
\begin{cases}
\frac{\partial f}{\partial x}+\lambda\frac{\partial g}{\partial x}
\;\equiv\; 2(x-x_0)+\lambda A \;=\; 0,\\
\frac{\partial f}{\partial y}+\lambda\frac{\partial g}{\partial y}
\;\equiv\; 2(y-y_0)+\lambda B \;=\; 0,\\
\frac{\partial f}{\partial z}+\lambda\frac{\partial g}{\partial z}
\;\equiv\; 2(z-z_0)+\lambda C \;=\; 0.
\end{cases}
\end{align}
We solve from (3) the differences
$$x-x_0 \;=\; -\frac{A\lambda}{2}, \quad y-y_0 \;=\; -\frac{B\lambda}{2}, \quad z-z_0 \;=\; -\frac{C\lambda}{2}$$
and set them into (1).\, It then yields the value
$$\lambda \;=\; \frac{2(D-E)}{A^2+B^2+C^2}$$
of the Lagrange multiplier, which we substitute into the preceding three equations obtaining
$$x-x_0 \;=\; \frac{A(D-E)}{A^2+B^2+C^2}, \quad y-y_0 \;=\; \frac{B(D-E)}{A^2+B^2+C^2},
\quad z-z_0 \;=\; \frac{C(D-E)}{A^2+B^2+C^2}.$$
These values give the minimal distance when put into the expression of $\sqrt{f}$:
$$d \;=\; \sqrt{\frac{(D-E)^2(A^2+B^2+C^2)}{(A^2+B^2+C^2)^2}}.$$
Hence we have gotten the distance \PMlinkescapetext{formula}
$$d \;=\; \frac{|D-E|}{\sqrt{A^2+B^2+C^2}}.$$