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

[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:

$$f(x)=ax^2+b$$

we obtain the 2X2 linear system:

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

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

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:

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

and for the logarithmic model:

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

 

[needalgebra]

whaaaaat how did you get that?

 

[Deveno]

Substituting in for [math]x[/math] obviously shows MarkFL's answer is correct...however, it does not directly give you [math]a[/math] and [math]b[/math].

To do this, re-write:

[math]f(x) = 30 + \frac{30}{\ln\left(\frac{12}{5}\right)}\ln\left( \frac{x}{25}\right)[/math]

[math] = 30 + \frac{30}{\ln\left(\frac{12}{5}\right)}\ln(x) - \frac{30\ln(25)}{\ln\left(\frac{12}{5}\right)}[/math]

[math]= 30\left(1 - \frac{\ln(25)}{\ln\left( \frac{12}{5}\right)}\right) + \frac{30}{\ln\left(\frac{12}{5}\right)}\ln(x)[/math]

which tells you:

[math]a = 30\left(1 - \frac{\ln(25)}{\ln\left( \frac{12}{5}\right)}\right) \approx -80.302[/math]

[math]b = \frac{30}{\ln\left(\frac{12}{5}\right)} \approx 34.267[/math]**********

Solving the linear system:

[math]a + b\ln(25) = 30[/math]
[math]a + b\ln(60) = 60[/math], I obtain (subtracting the top equation from the bottom):

[math]b(\ln(60) - \ln(25)) = 30[/math]

[math]b = \frac{30}{\ln(60) - \ln(25)} = \frac{30}{\ln\left( \frac{12}{5}\right)}[/math]

Substituting this in the first equation:

[math]a = 30 - b\ln(25) = 30 - \left( \frac{30}{\ln\left( \frac{12}{5}\right)} \right)\ln(25)[/math]

[math] = 30\left(1 - \frac{\ln(25)}{\ln\left( \frac{12}{5}\right)}\right) [/math]

which also agrees with MarkFL's answer.

 

[needalgebra]

ohhh ok.

I appreciate all of the help I am getting, thanks guys. I am actually done with this topic, i just want to know everything exactly how it is.

Thank you! How would i put that into wolfram for the graph?

and how would i do it without the included gigabytes, and the graph for that one?

 

[needalgebra]

any help guys please? :)