Chapter 11.5, Problem 40E

(a)

To determine

To find: the model of the form using graphing utility.

Answer to Problem 40E

a model based on this set of data can be given by these values:

  $a\approx -4.1\ast {10}^{-5}b\approx 1.62\ast {10}^{-2}c\approx -2.67d\approx 452.93$

Explanation of Solution

Given:

  $y=a{x}^3+b{x}^{2}cx+d$

Calculation:

First, input the $\left(x,y\right)$ pairs given in the table into a graphing calculator, and use the "regression" feature to find a model in the form $y=a{x}^3+b{x}^{2}cx+d$ . Doing so, find that a model based on this set of data can be given by these values:

  $a\approx -4.1\ast {10}^{-5}b\approx 1.62\ast {10}^{-2}c\approx -2.67d\approx 452.93$

Conclusion:

Therefore,a model based on this set of data can be given by these values:

  $a\approx -4.1\ast {10}^{-5}b\approx 1.62\ast {10}^{-2}c\approx -2.67d\approx 452.93$

(b)

To determine

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

Answer to Problem 40E

The graph was plotted.

Explanation of Solution

Calculation:

Now, plot the given data and the graph of our model in the same viewing window. The result should look like:

  

Conclusion:

Therefore,the graph was plotted.

(c)

To determine

To estimate: the area of the plot using model.

Answer to Problem 40E

The area pf the plot is $A\approx 78,504f{t}^2$ .

Explanation of Solution

Given:

Calculation:

Part C:

The limit of a fraction of polynomials of the same degree as $n\to \infty$ is the ratio of the coefficients of the terms of the highest degree:

  $\underset{n\to \infty }{\lim }A\left(n\right)=\frac{{300}^{4}a}{4}+\frac{2\ast {300}^{3}b}{6}+45000c+300d$

Substituting the values of a, b, c, and d found in part (a),

  $\begin{array}{l}A\approx \frac{{300}^4\left(-4.1\ast {10}^{-5}\right)}{4}+\frac{2\ast {300}^3\left(1.62\ast {10}^{-2}\right)}{6}+45000\left(-2.67\right)+300\left(452.93\right)\\ =78,504f{t}^2\end{array}$

Conclusion:

Therefore,the area pf the plot is $A\approx 78,504f{t}^2$ .