Chapter 12.5, Problem 46E

a.

To determine

To find: Use the regression feature of a graphing utility, to find a model of the form $y=a{x}^3+b{x}^2+cx+d$ .

Answer to Problem 46E

Buy using the regression feature of $y=\left(-4\times {10}^{-5}\right)x^3+0.06x^2-2.67x+453$ .

Explanation of Solution

Given:$y=a{x}^3+b{x}^2+cx+d$

x	0	50	100	`150	200	250	300
y	450	362	305	268	245	156	0

Calculation:

For the given values,

  

  $y=\left(-4\times {10}^{-5}\right)x^3+0.06x^2-2.67x+453$ this is the equation in the form of $y=a{x}^3+b{x}^2+cx+d$ found using the regression feature in a graphing utility.

b.

To determine

To plot: The data and graph the model in the same viewing window.

Answer to Problem 46E

  

Explanation of Solution

Given:$y=a{x}^3+b{x}^2+cx+d$

x	0	50	100	`150	200	250	300
y	450	362	305	268	245	156	0

Calculation:

For the given values,

x	0	50	100	`150	200	250	300
y	450	362	305	268	245	156	0

  

The above graph represents the data given in a viewing window.

c.

To determine

To find: The area of the model in part ‘a’.

Answer to Problem 46E

Area for the model in part ‘a’ is approximately 78750 square feet.

Explanation of Solution

Given:$y=a{x}^3+b{x}^2+cx+d$

x	0	50	100	`150	200	250	300
y	450	362	305	268	245	156	0

Calculation:

From the graph of part ‘b’,

  $\begin{array}{l}a=0\\ b\approx 300\end{array}$

Width of rectangle is $\frac{b-a}{n}=\frac{300}{n}$

Height of rectangle is $f\left(a+\frac{\left(b-a\right)}{n}\right)=f\left(\frac{300i}{n}\right)$

  $\begin{array}{l}A\approx {\displaystyle \sum _{i=1}^{n}f\left(\frac{300i}{n}\right)\frac{300}{n}}\\ A\approx {\displaystyle \sum _{i=1}^n\left[\left(-4\times {10}^{-5}\right){\left(\frac{300i}{n}\right)}^3+0.016{\left(\frac{300i}{n}\right)}^2-2.76\left(\frac{300i}{n}\right)+453\right]}\frac{300}{n}\\ A\approx {\displaystyle \sum _{i=1}^n\left[-\frac{324000i^3}{n^4}+\frac{432000i^2}{n^3}-\frac{240300i}{n^2}+\frac{135900}{n}\right]}\\ A\approx -\frac{324000}{n^4}{\displaystyle \sum _{i=1}^n{i}^3}+\frac{432000}{n^3}{\displaystyle \sum _{i=1}^n{i}^2}-\frac{240300}{n^2}{\displaystyle \sum _{i=1}^{n}i}+\frac{135900}{n}{\displaystyle \sum _{i=1}^{n}1}\\ \text{By using}{\displaystyle \sum _{i=1}^n{i}^3}=\frac{n^2{\left(n+1\right)}^2}{4}\text{,}{\displaystyle \sum _{i=1}^n{i}^2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\text{,}{\displaystyle \sum _{i=1}^{n}i}=\frac{n\left(n+1\right)}{2}\text{and}{\displaystyle \sum _{i=1}^{n}c}=cn\\ A\approx -\frac{81000{\left(n+1\right)}^2}{n^2}+\frac{72000\left(n+1\right)\left(2n+1\right)}{n^2}-\frac{120150\left(n+1\right)}{n}+135900\end{array}$

The Area is,

  $\begin{array}{l}A=\underset{x\to \infty }{\lim }\left[-\frac{81000{\left(n+1\right)}^2}{n^2}+\frac{72000\left(n+1\right)\left(2n+1\right)}{n^2}-\frac{120150\left(n+1\right)}{n}+135900\right]\\ A=\underset{x\to \infty }{\lim }\left[-81000-0-0+144000+0+0-120150-0+135900\right]\\ A\approx 78750\end{array}$

Thus we can estimate the area of the plot of land.