20.5: H1.05- Example 3 Alternative Method
- Page ID
- 51706
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Alternative method of solution for Example 3:
Example 3 question: For a certain type of letter sent by Federal Express, the charge is $8.50 for the first 8 ounces and $0.90 for each additional ounce (up to 16 ounces.) How much will it cost to send a 12-ounce letter?
We saw here that it is awkward for the value to be so far from any useful values for the variables. This suggests that we re-define the variables to be more useful.
1-2. Define the variables and their units and possible values. List the points.
Let x = number of ounces above 8. So the possible values for x are 0 to 8 (which corresponds to the letter weighing 8 ounces to 16 ounces.)
Let y = cost. The possible values for y are $8.50 and larger.
A letter weighing 8 ounces costs $8.50. So x = 0. The point is (0, 8.50)
A letter weighing 9 ounces costs $8.50+0.90 = $9.40. So x = 1. The point is (1, 9.40)
3. Is a linear model appropriate?
Yes, because the cost increases by the same amount for each additional ounce.
4-5. Find the slope and the formula.
Using the points above, we have
6. Interpret the y-intercept.
When x = 0, the value for y is 8.50. Since x = weight – 8 ounces, that means that when the letter weighs 8 ounces, the cost is $8.50.
Notice that defining the x-variable in this manner has made the y-intercept more meaningful in the problem.
7. Make the prediction.
If the letter weighs 12 ounces, then we use x = 12 – 8 = 4. So .
This agrees with the answer we found by the first method.
8. Graph and check whether you obtain the same results.
| weight | x = weight – 8 | y=cost |
| 8 | 0 | 8.5 |
| 9 | 1 | 9.4 |
By hand, plot these points, draw the line, extend it to the limits for the variables, and use it to estimate the answer to the question.
On the graph, look up a weight of 12 ounces, which is
x = 12 – 8 = 4.
The cost for x = 4 is a little larger than $12, which is consistent with the value we obtained from the formula.
- Mathematics for Modeling. Authored by: Mary Parker and Hunter Ellinger. License: CC BY: Attribution