How do I accurately model and calculate costs for different data plans?

[needalgebra]

Re: Are my answers correct? Quadratic and logarithmic modeling

I actually have a few more things that i'd like you to check for me please.

Answers:

Appetizer: {0<=x<=10 y=7 x>10 y=2(x-10)+7 or y=2x-13

Quest: {0<=x<=40 y=40 x>40 y=1(x-40)+40 or y=x)

Voyager: {0<=x<=60 y=60 x>60 y=.5(x-60)+60 or y=.5x+30

Question:

CURRENT PACKS:

develop equations that model the three current packs. including the domain for each equation.
find the intersection points of at least 3 different models.

appetizer : monthly access fee \$7, included gigabyte 10, cost per additional gigabyte \$2

quest: monthly access fee \$40, included gigabyte 40, cost per additional gigabyte \$1

voyager: monthly access fee \$60, included gigabyte 60, cost per additional gigabyte
\$0.50

Extra notes:

i don't exactly remember how to get the intersection points, i know its obviously when y crosses with x, but how would i graph this?

 

[MarkFL]

Yes, using linear models, you have correctly found:

Appetizer:

\(\displaystyle A(x)=\begin{cases}7 & 0\le x\le10\\ 2x-13 & 10<x \\ \end{cases}\)

Quest:

\(\displaystyle Q(x)=\begin{cases}40 & 0\le x\le40\\ x & 40<x \\ \end{cases}\)

Voyager:

\(\displaystyle V(x)=\begin{cases}60 & 0\le x\le60\\ \frac{1}{2}x+30 & 60<x \\ \end{cases}\)

A popular program online for plotting functions is Wolfram|Alpha: Computational Knowledge Engine.

Entering this command there:

piecewise[{{7,0<=x<=10},{2x-13,10<x}}],piecewise[{{40,0<=x<=40},{x,40<x}}],piecewise[{{60,0<=x<=60},{x/2+30,60<x}}] where x=0 to 100

produces this graph:

View attachment 1166

This will give you an idea where to look for the intersection points. To find them, set the two functions equal to one another (using the appropriate piece), and then solve for $x$, then use this value of $x$ in either function to find the value of $y$.

 

[needalgebra]

Thanks Mark! :D

I've got the same plotting question for my other post, how would i graph the logarithm and quadratic models?

 

[needalgebra]

I've got one more last question regarding this!

how do i model the effects on the cost to the customer by removing the "included gigabytes usage".

Is it the same as the graph you showed me but instead of having a horizontal line for the "included gigabytes" i just go straight down normally?

 

[MarkFL]

If you remove the included gigabytes, then this has the effect of shifting the graphs of the respective packages to the left by an amount so as to remove the horizontal portion of each graph. Thus, you would have:

\(\displaystyle A(x)=2x+7\)

\(\displaystyle Q(x)=x+40\)

\(\displaystyle V(x)=\frac{1}{2}x+60\)

 

[needalgebra]

oh ok and how would i put that on W.A?

 

[MarkFL]

oh ok and how would i put that on W.A?

I would use:

y=2x+7,y=x+40,y=x/2+60 where x=0 to 50

It's a lot simpler when the functions are not piecewise defined. :D

 

[needalgebra]

I can't get it to work...

 

[MarkFL]

What happens when you enter the above command?

 

[needalgebra]

never mind! it was just my browser! haha thanks! :D

 

[needalgebra]

ok so for my intersection points i got:

appetizer / quest : 33 , 73

appetizer / voyager : 26.5 , 60

Quest / voyager : 20 , 60

 

[MarkFL]

The first point is correct. Show what you did for the other two so we can figure out where you went wrong.

 

[needalgebra]

Quest & Voyager:

1x + 40 = + 0x + 60
1x - 0x = 60 - 40
+ 1x = 20
x = 20/ + 1
x = 20y = 1 * (20) + 40
y = 20 + 40
y = 60

Appetizer & Voyager:

2x + 7 = + 0x + 60
2x - 0x = 60 - 7
+ 2x = 53
x = 53/ + 2
x = 26.5

y = 2 * (26.5) + 7
y = 53 + 7
y = 60

 

[MarkFL]

Why are you using 0x + 60 for Voyager? You want x/2 + 60.

 

[needalgebra]

its the same thing.. i think. please do lead me the right way

 

[MarkFL]

They are not the same...1/2 times $x$ is different from 0 times $x$, in the same way that 1/2 is not equal to 0. Try solving the systems using x/2 + 60 for the Voyager model.

 

[needalgebra]

Appetizer & Voyager:

2x + 7 = + 0.5x + 60
2x - 0.5x = 60 - 7
+ 1.5x = 53
x = 53/ + 1.5
x = 35.3333333333y = 2 * (35.3333333333) + 7
y = 70.6666666667 + 7
y = 77.6667

 

[MarkFL]

Those are correct, well at least for decimal approximations. :D

Good work! You should have no trouble finding the remaining point.

 

[needalgebra]

woooooooooooooo! (Music)

(i forgot to mention that the three intersect points i just did were for the graphs without the included gigabytes)

is this one fine?

Quest & voyager
x + 40 = 0.5x + 60
x - 0.5x = 60 - 40
0.5x = 20
x = 20/ 0.5
x = 40

y = 1 * (40) + 40
y = 40 + 40
y = 80(this one right here is for the normal graph with the included gigabytes) is it fine?

Appetizer and Voyager:

2x - 13 = 0.5x + 30
2x - 0.5x = 30 - -13
1.5x = 43
x = 43/ + 1.5
x = 28.6666666667

y = 2 * (28.6666666667) - 13
y = 57.3333333333 - 13
y = 44.3333

How do i get the intersection points of the logarithm and quadratic models? (i think i should ask this on the other post)

 

[MarkFL]

woooooooooooooo! (Music)

(i forgot to mention that the three intersect points i just did were for the graphs without the included gigabytes)

is this one fine?

Quest & voyager
x + 40 = 0.5x + 60
x - 0.5x = 60 - 40
0.5x = 20
x = 20/ 0.5
x = 40

y = 1 * (40) + 40
y = 40 + 40
y = 80

Yes, that is correct. You can check by substitution:

\(\displaystyle 40+40=\frac{40}{2}+60\)

\(\displaystyle 80=80\)

This is true, so you know your solution is correct.

(this one right here is for the normal graph with the included gigabytes) is it fine?

Appetizer and Voyager:

2x - 13 = 0.5x + 30
2x - 0.5x = 30 - -13
1.5x = 43
x = 43/ + 1.5
x = 28.6666666667

y = 2 * (28.6666666667) - 13
y = 57.3333333333 - 13
y = 44.3333

If you notice from the graph, the Appetizer line intersects the other piece of the Voyager line, i.e., the horizontal part.

 

[needalgebra]

so how would i get that?

 

[MarkFL]

Equate the non-horizontal portion of Appetizer to the horizontal portion of Voyager.

 

[needalgebra]

i have to find the intersection points for appetizer & quest, appetizer & voyager and quest & voyager.

Im not exactly understanding what you mean.

i mean for example, with appetizer and voyager, what would happen to 0 - 10 on the x-axis and the rest after 60 for voyager..

im kind of confused.

This is for the normal graph, with the included gigabytes. i already figured out the other one

 

[MarkFL]

Using the graph as a guide, we can see that it is the non-horizontal portion of Appetizer that intersects with the horizontal portion of Voyager. So, you already know the $y$-coordinate of this point, you simply need to solve the resulting equation for $x$ to get that coordinate.

 

[needalgebra]

ugggh. i just saw that i have to model the effects to the customer of removing the "included gigabytes usage". i guess it means for these including the logarithmic and quadratic functions... how do i do what the question is asking me?

 

[MarkFL]

Can you give the problem here, in its entirety, as it is given to you?

 

[needalgebra]

The question says:

model the effects on the cost to the customer of removing the "included gigabytes usage'.

 

[MarkFL]

Isn't that what we've done?

I was hoping you would present the problem in its entirety, from beginning to end, including everything.

 

[needalgebra]

Yeah, sure i can do that no problem!

an internet provider is reviewing its monthly broadband usage packs, currently, customers can choose one of the three following options:

CURRENT PACKS:
appetizer : monthly access fee \$7, included gigabyte 10, cost per additional gigabyte \$2
quest: monthly access fee \$40, included gigabyte 40, cost per additional gigabyte \$1
voyager: monthly access fee \$60, included gigabyte 60, cost per additional gigabyte \$0.50

these packs comprise a monthly access fee which includes usage up to a certain number of gigabytes plus an additional cost per gigabyte over and above that.

The internet provider intends to discontinue the quest pack. they are considering two new pack options to replace it.

Replacement pack options:

(option 1) quadratic model, monthly access fee \$20, included gigabytes 20, cost per additional gigabyte *see below (*)

(option 2) logarithmic model, monthly access fee \$30, included gigabytes 25, cost per additional gigabyte *see below (**)

(*) The charge option 1 will increase quadratically after 20 gigabytes with customers paying the same amount as voyager customers at 60 gigabytes.

(**) The charge option 2 will increase logarithmically in the form ƒ(x) = a + b 1n|x| after 25 gigabytes with customers also paying the same amount as voyager customers at 60 gigabytes.

in addition, they are also looking into the possibility of removing the 'included gigabyte usage' from each pack . this would mean that the customer would pay the monthly access fee plus the cost of each gigabyte used.

this activity requires you to write a report to the internet provider including
(i) a recommendation of an appropriate new pricing pack for medium users to replace the quest pack.
(ii) an analysis of the effect of the cost to the customer of removing the 'included gigabytes usage' from your recommended replacement of the quest pack option and the two remaining packs, appetizer and voyager.

Task:

-graph the current pack options -appetizer, quest and voyager.
-graph the two replacement pack options - option 1 - quadratic model, option 2 - logarithmic model.
-develop equations that model the three current and the two replacement pack options, including the domain for each equation - full working must be shown.
-find the intersection points of at least 3 different models.
-model the effects on the cost to the customer of removing the 'included gigabytes usage'



when analyzing the results of your modeling you will need to:

compare and contrast the different plans, describing the positive and negative aspects of each plan for potential customers

recommend which of the replacement pack options should replace the quest pack option, with justification

state any assumptions, generalizations and reflections you make during your modelling

discuss the effect on the cost to the customer of removing the "included gigabyte usage"

provide clear recommendations to the internet provider based on your findings

 

[MarkFL]

Okay, I suspected something was omitted from the original statement of the problem, which I have highlighted in red:

(*) The charge option 1 will increase quadratically after 20 gigabytes with customers paying the same amount as voyager customers at 60 gigabytes.

(**) The charge option 2 will increase logarithmically in the form ƒ(x) = a + b 1n|x| after 25 gigabytes with customers also paying the same amount as voyager customers at 60 gigabytes.

This changes things, and means we need to start over with the quadratic and logarithmic models. Can you state the two points through which each model must pass?

 

[needalgebra]

eehh. what do you mean?

 

[MarkFL]

We are implicitly given two points through which both curves must pass.

For example, for the quadratic model, we know the quadratic portion of the function begins at (20,20), but must also pass through the point (60,60), since this is the amount for voyager customers at 60 gigabytes.

So, using:

\(\displaystyle f(x)=ax^2+b\)

we obtain the 2X2 linear system:

\(\displaystyle f(20)=a(20)^2+b=400a+b=20\)

\(\displaystyle f(60)=a(60)^2+b=3600a+b=60\)

Now you may determine the values of the parameters $a$ and $b$.

Then follow a similar process for the logarithmic model.

 

[needalgebra]

(quadratic model):

f(x)=ax2+b

we obtain the 2X2 linear system:

where:

f(20)=a(20)2+b=400a+b=20

f(60)=a(60)2+b=3600a+b=60

Logarithmic model:

f(x) = a + b Ln(x)

where:

f(25) = a + b (25)Ln = 30

f(60) = a + b (60)Ln =60answer:

quadratic model:

f(60)=60, 3600a+(20-400a)=60, 3200a=40 and a=4/320=1/80.
b=20-400/80=20-5=15.

logarithmic model:

f(25)=30, a=30-bLn(25), f(60)=60, 30-bLn(60)=60, b=-30/Ln(60)=-7.33, a=30-(-30/Ln(60)*Ln(25)) =30(1+Ln(25)/Ln(60))=53.6

 

[needalgebra]

ehh is that right?

 

[MarkFL]

For the quadratic model, I get:

\(\displaystyle f(x)=\frac{1}{80}x^2+15\)

and for the logarithmic model:

\(\displaystyle f(x)=30\left(1+\frac{\ln\left(\frac{x}{25} \right)}{\ln\left(\frac{12}{5} \right)} \right)\)