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Mathematical modeling

needalgebra

Member
Aug 13, 2013
45
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 dont exactly remember how to get the intersection points, i know its obviously when y crosses with x, but how would i graph this?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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:

needalgebra.jpg

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

Member
Aug 13, 2013
45
Thanks Mark! :D

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

needalgebra

Member
Aug 13, 2013
45
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Member
Aug 13, 2013
45
oh ok and how would i put that on W.A?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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

Member
Aug 13, 2013
45
I cant get it to work....
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
What happens when you enter the above command?
 

needalgebra

Member
Aug 13, 2013
45
never mind! it was just my browser! haha thanks! :D
 

needalgebra

Member
Aug 13, 2013
45
ok so for my intersection points i got:

appetizer / quest : 33 , 73

appetizer / voyager : 26.5 , 60

Quest / voyager : 20 , 60
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The first point is correct. Show what you did for the other two so we can figure out where you went wrong.
 

needalgebra

Member
Aug 13, 2013
45
Quest & Voyager:

1x + 40 = + 0x + 60
1x - 0x = 60 - 40
+ 1x = 20
x = 20/ + 1
x = 20


y = 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

Administrator
Staff member
Feb 24, 2012
13,775
Why are you using 0x + 60 for Voyager? You want x/2 + 60.
 

needalgebra

Member
Aug 13, 2013
45
its the same thing.. i think. please do lead me the right way
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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

Member
Aug 13, 2013
45
Appetizer & Voyager:

2x + 7 = + 0.5x + 60
2x - 0.5x = 60 - 7
+ 1.5x = 53
x = 53/ + 1.5
x = 35.3333333333


y = 2 * (35.3333333333) + 7
y = 70.6666666667 + 7
y = 77.6667
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Those are correct, well at least for decimal approximations. :D

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

needalgebra

Member
Aug 13, 2013
45
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Member
Aug 13, 2013
45
so how would i get that?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Equate the non-horizontal portion of Appetizer to the horizontal portion of Voyager.
 

needalgebra

Member
Aug 13, 2013
45
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Member
Aug 13, 2013
45
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?