# Math Help - Mathematical Modelling-Problem

1. ## Mathematical Modelling-Problem

Hello!
I need to solve this problem and I am not sure if I did it right.

"The value of an automobile over time is given in the following table:
1 | 14000
2 | 9100
3 | 6200
4 | 4000
5 | 3000

a) Use graphing software or an algebraic method to determine an equation that fits
i) the linear model
iii) the exponential model
b) Use each model to predict the value of the car after 10 years

d) Which function provides the best model? "
***
a) Linear: y=mx+b
slope-m, y-intercept(b)

m=(y2-y1)/(x2-x1)=(3000-14000)/(5-1)= (-11000)/(4)=-2750
point(1,14000) -> y=-2750(x-1)+14000
y=-2750x+16750
p and q-coordinates of the vertex
(1,14000) is the highest point on the graph -> (p,q)
equation: y=a(x-1)^2+14000
(5,3000) for x,y
3000=a(5-1)^2+14000
3000=a*16+14000
-11000=16a
a=-687.5

Exponential: y=ar^x, where r is the constant ratio
Ratio: 9100/14000=0.65 ; 6200/9100=0.68 ; 4000/6200=0.65 ; 3000/4000=0.75
Average: 0.6825 ---> r=0.68
y=a(0.68)^x with y=14000 when x=1 ---> 14000=a(0.68)^1 or a=14000/0.68=20.59
The exponential equation is: y=21(0.68)^x

b) After 10 years ...Linear:y= -10750 , Quadratic: y= -41687.5 , Exponential: y=0.44
c)Exponential is the most reasonable based upon the 10th year values.

Did I solved it correct? Are there any mistakes?

Thank you.
Laura

2. ## Re: Mathematical Modelling-Problem

Originally Posted by Laura05
Hello!
I need to solve this problem and I am not sure if I did it right.

"The value of an automobile over time is given in the following table:
1 | 14000
2 | 9100
3 | 6200
4 | 4000
5 | 3000

a) Use graphing software or an algebraic method to determine an equation that fits
i) the linear model
iii) the exponential model
b) Use each model to predict the value of the car after 10 years

d) Which function provides the best model? "
***
... all your considerations and calculations seem OK for me - with one exception:
y=a(0.68)^x with y=14000 when x=1 ---> 14000=a(0.68)^1 or a=14000/0.68=20.59
The exponential equation is: y=21(0.68)^x

b) After 10 years ...Linear:y= -10750 , Quadratic: y= -41687.5 , Exponential: y=0.44
c)Exponential is the most reasonable based upon the 10th year values.

Did I solved it correct? Are there any mistakes?

Thank you.
Laura
The calculation marked in red is definitely wrong:

$a=14000/0.68 \approx 20600$

Btw: I believe that you - even in Canada - cann't buy a ten year old car for 44 cents

3. ## Re: Mathematical Modelling-Problem

Maybe a 10 year old toy car ?

Thanks. Calculation error.
***
The exponential equation is: y=20588.24(0.68)^x

b) After 10 years ...
[...] Exponential: y=435.22
Now it's ok?!
*
Can you give me a little hit for d ? I think (maybe I'm wrong) that the best model is the exponential one? Points c and d seems similar to me.

4. ## Re: Mathematical Modelling-Problem

Originally Posted by Laura05
Maybe a 10 year old toy car ? This would be a toy from the antiques shop and then you probably have to pay quite a lot more then 44 cts.

Thanks. Calculation error.
***
The exponential equation is: y=20588.24(0.68)^x

b) After 10 years ...
[...] Exponential: y=435.22
Now it's ok?!
*
Can you give me a little hit for d ? I think (maybe I'm wrong) that the best model is the exponential one? Points c and d seems similar to me.
When I tried to give you some hints to answer d) I noticed that your quadratic model is wrong:

The general equation of a quadratic function is:

$y = ax^2+bx+c$

Use 3 of the "points" (value of car over time) and plug in the the year(for x) and the value(for y) to get a system of 3 simultaneous equations in (a,b,c). For instance:
$\begin{array}{rcl}14000&=&a+b+c \\ 9100&=&4a+2b+c \\ 3000&=&25a+5b+c \end{array}$

Solve for (a,b,c). You should come out with:

$y = 716.7 x^2-7050x+20333$

To do d) I would graph the "points" and the graphs of the 3 functions and would try to judge which function describes best the development of the car value over time.