Chapter 12.5, Problem 48E

a.

To determine

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 48E

  $\boxed{y=-4.0\times {10}^{-5}{x}^3+0.016x^2-2.64x+452.64}$

Explanation of Solution

Given information:

The table shows the measurements (in feet) of a lot bounded by a stream and two straight roads that meet at right angles (see figure).

  $\begin{array}{cccccccc}x& 0& 50& 100& 150& 200& 250& 300\\ y& 450& 362& 305& 268& 245& 156& 0\end{array}$

  

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

Calculation:

Consider the following table

  $\begin{array}{cccccccc}x& 0& 50& 100& 150& 200& 250& 300\\ y& 450& 362& 305& 268& 245& 156& 0\end{array}$

To determine the regresion model using the graphing utility $TI-83$ ,proceed as follows.

Press stat, the display will be

  

Now select edit, the display will be

  

Now provide the data under the columns of list 1 and list 2.

  

Now press stat and choose calc, the display will be

  

Now choose cubic reg. the display will be

  

Now press enter. The display will be

  

Hence the cubic regression model of the above data is

  $\boxed{y=-4.0\times {10}^{-5}{x}^3+0.016x^2-2.64x+452.64}$

b.

To determine

Plot a utility graph.

Answer to Problem 48E

  

Explanation of Solution

Given information:

The table shows the measurements (in feet) of a lot bounded by a stream and two straight roads that meet at right angles (see figure).

  $\begin{array}{cccccccc}x& 0& 50& 100& 150& 200& 250& 300\\ y& 450& 362& 305& 268& 245& 156& 0\end{array}$

  

Use the graphing utility to plot the data and graph the model in the same viewing window.

Calculation:

Consider the following table

  $\begin{array}{cccccccc}x& 0& 50& 100& 150& 200& 250& 300\\ y& 450& 362& 305& 268& 245& 156& 0\end{array}$

To plot the data, proceed as follows.

Press 2^nd Y. The display will be

  

Now press enter and choose on. The display will be

  

Now press Window to choose the proper scale. The display will be

  

Now press graph, the display will be

  

To plot the regression equation, proceed as follows.

Now press $Y$ and choose plot 2 using up arrow key. The display will be

  

Now provide the equation, the display will be

  

Now press graph, the display will be

  

c.

To determine

Find the area of the lot.

Answer to Problem 48E

  $\boxed{79992}$ Square feet

Explanation of Solution

Given information:

The table shows the measurements (in feet) of a lot bounded by a stream and two straight roads that meet at right angles (see figure).

  $\begin{array}{cccccccc}x& 0& 50& 100& 150& 200& 250& 300\\ y& 450& 362& 305& 268& 245& 156& 0\end{array}$

  

Use the model in part (a) to estimate the area of the lot.

Calculation:

Define the dimensions of the rectangles as follows:

Width:

  $\begin{array}{l}\frac{b-a}{n}=\frac{300-0}{n}\\ =\frac{300}{n}\end{array}$

Height:

  $\begin{array}{l}f\left(a+\frac{\left(b-a\right)i}{n}\right)=f\left(0+\frac{300i}{n}\right)\\ =f\left(\frac{300i}{n}\right)\\ f\left(x\right)=-4.0\times {10}^{-5}{x}^3+0.016x^2-2.64x+452.64\end{array}$

Hence,

  $f\left(\frac{300i}{n}\right)=-4.0\times {10}^{-5}{\left(\frac{300i}{n}\right)}^3+0.016{\left(\frac{300i}{n}\right)}^2-2.64\left(\frac{300i}{n}\right)+452.64$

  $=-1080\frac{i^3}{n^3}+1440\frac{i^2}{n^2}-792\frac{i}{n}+452.64$

Now approximate the area as the sum of the areas of $n$ rectangles.

  $\begin{array}{l}A\approx {\displaystyle \sum _{i=1}^{n}f}\left(a+\frac{b-a}{n}\right)\left(\frac{b-a}{n}\right)\\ \approx {\displaystyle \sum _{i=1}^n\left(-1080\frac{i^3}{n^3}+1440\frac{i^2}{n^2}-792\frac{i}{n}+452.64\right)}\left(\frac{300}{n}\right)\\ \approx 300{\displaystyle \sum _{i=1}^n\left(-1080\frac{i^3}{n^4}+1440\frac{i^2}{n^3}-792\frac{i}{n^2}+\frac{452.64}{n}\right)}\\ \approx -\frac{324000}{n^4}{\displaystyle \sum _{i=1}^n{i}^3}+\frac{432000}{n^3}{\displaystyle \sum _{i=1}^n{i}^2}-\frac{237600}{n^2}{\displaystyle \sum _{i=1}^{n}i}+\frac{135792}{n}{\displaystyle \sum _{i=1}^{n}1}\end{array}$

Now use the properties of summations.

  $\begin{array}{l}{\displaystyle \sum _{i=1}^n{i}^3}={\left[\frac{n\left(n+1\right)}{2}\right]}^2\\ {\displaystyle \sum _{i=1}^n{i}^2}=\left[\frac{n(n+1)(2n+1)}{6}\right]\\ {\displaystyle \sum _{i=1}^{n}i}=\left[\frac{n(n+1)}{2}\right]\\ {\displaystyle \sum _{i=1}^{n}1}=n\end{array}$

Now find the exact area by taking the limit as $n$ approaches.

  $A=\underset{x\to \infty }{\lim }\left\{-81000{\left[\frac{n+1}{n}\right]}^2+72000\left[\frac{\left(n+1\right)\left(2n+1\right)}{n^2}\right]-118800\left[\frac{(n+1)}{n}\right]+135792\right\}$

Divide individually within the paranthesis

  $A=\underset{x\to \infty }{\lim }-81000{\left[1+\frac{1}{n}\right]}^2+72000\left[\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)\right]-118800\left[1+\frac{1}{n}\right]+135792$

Now replace $\frac{1}{n}$ by $0$

  $\begin{array}{l}A=-81000+144000-118800+135792\\ =79992\end{array}$

Hence, the under the curve is $\boxed{79992}$ sqaure feet.