Q 13. Minima and Maxima - Calculus

Question

Show that the unique solution of the system of equations 

$$\begin{gathered} 2\left(x-x_0\right)-2·\frac{a}{c}\left(\frac{d-ax-by}{c}-z_0\right)=0 \\ 2\left(y-y_0\right)-2·\frac{b}{c}\left(\frac{d-ax-by}{c}-z_0\right)=0 \end{gathered}$$

is $\left(\frac{ad-ab{y}_0-ac{z}_0+b^2{x}_0+c^2{x}_0}{a^2+b^2+c^2},\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}\right)$

Step-by-Step Solution

Answer

The solution of the system of equations is $\left(\frac{ad-ab{y}_0-ac{z}_0+b^2{x}_0+c^2{x}_0}{a^2+b^2+c^2},\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}\right)$

Step 1: Given information

The system of equations $$\begin{gathered} 2\left(x-x_0\right)-2·\frac{a}{c}\left(\frac{d-ax-by}{c}-z_0\right)=0 \\ 2\left(y-y_0\right)-2·\frac{b}{c}\left(\frac{d-ax-by}{c}-z_0\right)=0 \end{gathered}$$

Step 2: Calculation

Consider the system of equation

$$\begin{gathered} 2\left(x-x_0\right)-2·\frac{a}{c}\left(\frac{d-ax-by}{c}-z_0\right)=0\dots \dots \left(1\right) \\ 2\left(y-y_0\right)-2·\frac{b}{c}\left(\frac{d-ax-by}{c}-z_0\right)=0\dots \dots \left(2\right) \end{gathered}$$

The goal is to figure out how to solve the system of equations.

Rewrite equation $\left(1\right)$ as

$$\begin{gathered} 2\left(x-x_0\right)-2·\frac{a}{c}\left(\frac{d-ax-by}{c}-z_0\right)=0 \\ 2x-2x_0-\frac{2ad}{c^2}+\frac{2a^2}{c^2}x+\frac{2aby}{c^2}+\frac{2a}{c}{z}_0=0 \\ 2x+\frac{2a^2}{c^2}x-2x_0-\frac{2ad}{c^2}+\frac{2aby}{c^2}+\frac{2a}{c}{z}_0=0 \\ \left(2+\frac{2a^2}{c^2}\right)x+\frac{2ab}{c^2}y-2x_0-\frac{2ad}{c^2}+\frac{2a}{c}{z}_0=0 \\ \left(2+\frac{2a^2}{c^2}\right)x+\frac{2ab}{c^2}y=2x_0+\frac{2ad}{c^2}-\frac{2a}{c}{z}_0\dots \dots \end{gathered}$$

Rewrite equation ( 2 ) as

$\begin{array}{r}2\left(y-y_0\right)-2·\frac{b}{c}\left(\frac{d-ax-by}{c}-z_0\right)=0\\ 2y-2y_0-\frac{2bd}{c^2}+\frac{2ab}{c^2}x+\frac{2b^{2}y}{c^2}+\frac{2b}{c}{z}_0=0\\ 2y+\frac{2b^2}{c^2}y-2y_0-\frac{2bd}{c^2}+\frac{2ab}{c^2}x+\frac{2b}{c}{z}_0=0\\ \left(2+\frac{2b^2}{c^2}\right)y+\frac{2ab}{c^2}x-2y_0-\frac{2bd}{c^2}+\frac{2b}{c}{z}_0=0\\ \frac{2ab}{c^2}x+\left(2+\frac{2b^2}{c^2}\right)y=2y_0+\frac{2bd}{c^2}-\frac{2b}{c}{z}_0\dots \dots \end{array}$

$$\begin{gathered} \text{ Multiply ( }\left.3\text{ ) by }\frac{2ab}{c^2}\text{ and multiply ( }4\right)\text{ by }\left(2+\frac{2a^2}{c^2}\right)\text{ and then subtract } \\ \frac{2ab}{c^2}\left\{\left(2+\frac{2a^2}{c^2}\right)x+\frac{2ab}{c^2}y\right\}-\left(2+\frac{2a^2}{c^2}\right)\left\{\frac{2ab}{c^2}x+\left(2+\frac{2b^2}{c^2}\right)y\right\} \\ =\frac{2ab}{c^2}\left(2x_0+\frac{2ad}{c^2}-\frac{2a}{c}{z}_0\right)-\left(2+\frac{2a^2}{c^2}\right)\left(2y_0+\frac{2bd}{c^2}-\frac{2b}{c}{z}_0\right) \end{gathered}$$

$$\begin{gathered} \frac{2ab}{c^2}\left(2+\frac{2a^2}{c^2}\right)x+\frac{2ab}{c^2}·\frac{2ab}{c^2}y-\left(2+\frac{2a^2}{c^2}\right)\frac{2ab}{c^2}x-\left(2+\frac{2a^2}{c^2}\right)\left(2+\frac{2b^2}{c^2}\right)y \\ =\frac{4ab}{c^2}{x}_0+\frac{4a^{2}bd}{c^4}-\frac{4a^{2}b}{c^3}{z}_0-4y_0-\frac{4bd}{c^2} \end{gathered}$$

$$\begin{gathered} +\frac{4b}{c}{z}_0-\frac{4a^2}{c^2}{y}_0-\frac{4a^{2}bd}{c^4}+\frac{4a^{2}b}{c^3}{z}_0 \\ \left\{\frac{4a^2{b}^2}{c^4}-\left(2+\frac{2a^2}{c^2}\right)\left(2+\frac{2b^2}{c^2}\right)\right\}y=\frac{4ab}{c^2}{x}_0-4y_0-\frac{4bd}{c^2}+\frac{4b}{c}{z}_0-\frac{4a^2}{c^2}{y}_0 \\ \left(\frac{4a^2{b}^2}{c^4}-4-\frac{4b^2}{c^2}-\frac{4a^2}{c^2}-\frac{4a^2{b}^2}{c^4}\right)y=-\frac{4bd}{c^2}+\frac{4ab}{c^2}{x}_0+\frac{4b}{c}{z}_0-4y_0-\frac{4a^2}{c^2}{y}_0 \end{gathered}$$

$$\begin{gathered} \left(-4-\frac{4b^2}{c^2}-\frac{4a^2}{c^2}\right)y=-\frac{4bd}{c^2}+\frac{4ab}{c^2}{x}_0+\frac{4b}{c}{z}_0-4y_0-\frac{4a^2}{c^2}{y}_0 \\ \left(\frac{-4c^2-4b^2-4a^2}{c^2}\right)y=\frac{-4bd+4ab{x}_0+4bc{z}_0-4c^2{y}_0-4a^2{y}_0}{c^2} \\ -\frac{4}{c^2}\left(a^2+b^2+c^2\right)y=-\frac{4}{c^2}\left(bd-ab{x}_0-bc{z}_0+c^2{y}_0+a^2{y}_0\right) \\ \left(a^2+b^2+c^2\right)y=bd-ab{x}_0-bc{z}_0+c^2{y}_0+a^2{y}_0 \\ y=\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2} \end{gathered}$$

Substitute $\text{ Substitute }y=\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}\text{ in ( 4) }$

$$\begin{gathered} \frac{2ab}{c^2}x+\left(2+\frac{2b^2}{c^2}\right)\left(\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}\right)=2y_0+\frac{2bd}{c^2}-\frac{2b}{c}{z}_0 \\ \frac{2}{c^2}\left\{abx+\left(c^2+b^2\right)\left(\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}\right)\right\}=\frac{2}{c^2}\left(c^2{y}_0+bd-bc{z}_0\right) \\ abx+\left(c^2+b^2\right)\left(\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}\right)=c^2{y}_0+bd-bc{z}_0 \end{gathered}$$

Step 3: Calculation

$abx=c^2{y}_0+bd-bc{z}_0-\left(c^2+b^2\right)\left(\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}\right)$

$=\frac{c^2{a}^2{y}_0+c^2\left(c^2+b^2\right)y_0+a^{2}bd+\left(c^2+b^2\right)bd-a^{2}bc{z}_0-\left(c^2+b^2\right)bc{z}_0}{a^2+b^2+c^2}$

$-\left(c^2+b^2\right)bd+\left(c^2+b^2\right)ab{x}_0+\left(c^2+b^2\right)bc{z}_0-\left(c^2+b^2\right)a^2{y}_0-c^2\left(c^2+b^2\right)y_0$

$=\frac{c^2{a}^2{y}_0+a^{2}bd-a^{2}bc{z}_0+\left(c^2+b^2\right)ab{x}_0-\left(c^2+b^2\right)a^2{y}_0}{a^2+b^2+c^2}$

$=\frac{c^2{a}^2{y}_0+a^{2}bd-a^{2}bc{z}_0+c^{2}ab{x}_0+b^{2}ab{x}_0-c^2{a}^2{y}_0-b^2{a}^2{y}_0}{a^2+b^2+c^2}$

$=\frac{a^{2}bd-a^{2}bc{z}_0+c^{2}ab{x}_0+b^{2}ab{x}_0-b^2{a}^2{y}_0}{a^2+b^2+c^2}$

$=ab\left(\frac{ad-ac{z}_0+c^2{x}_0+b^2{x}_0-ab{y}_0}{a^2+b^2+c^2}\right)$

$x=\frac{ad-ab{y}_0-ac{z}_0+b^2{x}_0+c^2{x}_0}{a^2+b^2+c^2}$

As a result, the system of equations solution is  

$\left(\frac{ad-ab{y}_0-ac{z}_0+b^2{x}_0+c^2{x}_0}{a^2+b^2+c^2},\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}\right)$