Chapter 7.3, Problem 110E

To determine

To calculate: The system of linear equations with the help of given data and then find the relationship between number of sides and combined number of sides and diagonals.

Answer to Problem 110E

Therefore the relationship is given by $y=\frac{1}{2}{x}^2-\frac{1}{2}x$ .

Explanation of Solution

Given information:

The number of sides and sides with diagonals are given as

  $\left(3,3\right),\left(4,6\right),\left(5,10\right)$ .

Formula used:

Use the below formulas for system of equations

  $\begin{array}{c}nc+\left({\displaystyle \sum _{i=1}^n{x}_i}\right)b+\left({\displaystyle \sum _{i=1}^n{x}_i^2}\right)a={\displaystyle \sum _{i=1}^n{y}_i}\\ \left({\displaystyle \sum _{i=1}^n{x}_i}\right)c+\left({\displaystyle \sum _{i=1}^n{x}_i^2}\right)b+\left({\displaystyle \sum _{i=1}^n{x}_i^3}\right)a={\displaystyle \sum _{i=1}^n{x}_i{y}_i}\\ \left({\displaystyle \sum _{i=1}^n{x}_i^2}\right)c+\left({\displaystyle \sum _{i=1}^n{x}_i^3}\right)b+\left({\displaystyle \sum _{i=1}^n{x}_i^4}\right)a={\displaystyle \sum _{i=1}^n{x}_i^2{y}_i}\end{array}$ .

Calculation:

As per given table the points are $\left(3,3\right),\left(4,6\right),\left(5,10\right)$ .

Now calculate

  $\begin{array}{c}{\displaystyle \sum _{i=1}^n{x}_i}=x_1+x_2+x_3\\ =3+4+5\\ =12\end{array}$ , $\begin{array}{c}{\displaystyle \sum _{i=1}^n{x}_i^2}=x_1^2+x_2^2+x_3^2\\ ={\left(3\right)}^2+{\left(4\right)}^2+{\left(5\right)}^2\\ =9+16+25\\ =50\end{array}$ , $\begin{array}{c}{\displaystyle \sum _{i=1}^n{x}_i^3}=x_1^3+x_2^3+x_3^3\\ ={\left(3\right)}^3+{\left(4\right)}^3+{\left(5\right)}^3\\ =27+64+125\\ =216\end{array}$

Also calculate

  $\begin{array}{c}{\displaystyle \sum _{i=1}^n{x}_i^4}=x_1^4+x_2^4+x_3^4\\ ={\left(3\right)}^4+{\left(4\right)}^4+{\left(5\right)}^4\\ =81+256+625\\ =962\end{array}$ , $\begin{array}{c}{\displaystyle \sum _{i=1}^n{y}_i}=y_1+y_2+y_3\\ =3+6+10\\ =19\end{array}$

Also calculate

  $\begin{array}{c}{\displaystyle \sum _{i=1}^n{x}_i{y}_i}=x_1{y}_1+x_2{y}_2+x_3{y}_3\\ =3\times 3+4\times 6+5\times 10\\ =9+24+50\\ =83\end{array}$ ,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^n{x}_i^2{y}_i}=x_1^2{y}_1+x_2^2{y}_2+x_3^2{y}_3\\ ={\left(3\right)}^2\times 3+{\left(4\right)}^2\times 6+{\left(5\right)}^2\times 10\\ =27+96+250\\ =373\end{array}$

Now substitute all the values in the formula

  $\begin{array}{c}3c+12b+50a=19\dots \left(1\right)\\ 12c+50b+216a=83\dots \left(2\right)\\ 50c+216b+962a=373\dots \left(3\right)\end{array}$

Solve equation $\left(1\right)$

  $\begin{array}{l}3c+12b+50a=19\\ 3c=19-50a-12b\\ c=\frac{19-50a-12b}{3}\end{array}$

Substitute the value of $c$ in equation $\left(2\right)$

  $\begin{array}{l}12c+50b+216a=83\\ 12\left(\frac{19-50a-12b}{3}\right)+50b+216a=83\\ 76-200a-48b+50b+216a=83\\ 16a+2b=7\dots \left(4\right)\end{array}$

Substitute the value of $c$ in equation $\left(3\right)$

  $\begin{array}{l}50c+216b+962a=373\\ 50\left(\frac{19-50a-12b}{3}\right)+216b+962a=373\\ 950-2500a-600b+648b+2886a=1119\\ 386a+48b=169\dots \left(5\right)\end{array}$

Solve equation $\left(4\right)$

  $\begin{array}{l}16a+2b=7\\ 2b=7-16a\\ b=\frac{7-16a}{2}\end{array}$

Now substitute the value of $b$ in equation $\left(5\right)$

  $\begin{array}{l}386a+48\left(\frac{7-16a}{2}\right)=169\\ 386a+168-384a=169\\ 2a=1\\ a=\frac{1}{2}\end{array}$

Substitute the value of $a$ in equation $\left(4\right)$

  $\begin{array}{l}16\left(\frac{1}{2}\right)+2b=7\\ 8+2b=7\\ 2b=-1\\ b=-\frac{1}{2}\end{array}$

Now substitute both the value of $b,c$ in equation $\left(1\right)$

  $\begin{array}{l}3c+12b+50a=19\\ 3c+12\left(-\frac{1}{2}\right)+50\left(\frac{1}{2}\right)=19\\ 3c-6+25=19\\ 3c=0\\ c=0\end{array}$ .

Therefore, the least square regression parabola

  $\begin{array}{l}y=a{x}^2+bx+c\\ y=\frac{1}{2}{x}^2-\frac{1}{2}x\end{array}$ .

Therefore, the equation of regression parabola is $y=\frac{1}{2}{x}^2-\frac{1}{2}x$ .