Results 1 to 4 of 4

Math Help - Mathematical Modelling-Problem

  1. #1
    Newbie
    Joined
    Jul 2013
    From
    Toronto
    Posts
    2

    Question 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
    ii) the quadratic model
    iii) the exponential model
    b) Use each model to predict the value of the car after 10 years

    c) Which result is most reasonable? Give reasons for your answer

    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
    Quadratic: y=a(x-p)^2 + q
    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
    The quadratic equation: y=-687.5(x-1)^2+14000

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123

    Re: Mathematical Modelling-Problem

    Quote Originally Posted by Laura05 View Post
    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
    ii) the quadratic model
    iii) the exponential model
    b) Use each model to predict the value of the car after 10 years

    c) Which result is most reasonable? Give reasons for your answer

    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2013
    From
    Toronto
    Posts
    2

    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123

    Re: Mathematical Modelling-Problem

    Quote Originally Posted by Laura05 View Post
    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.
    Attached Thumbnails Attached Thumbnails Mathematical Modelling-Problem-carprices.png  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. mathematical modelling!
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: June 20th 2013, 05:43 AM
  2. Really Struggling with Mathematical Modelling
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: April 15th 2011, 12:59 AM
  3. Mathematical Modelling through ODE
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: February 19th 2011, 08:04 PM
  4. Replies: 1
    Last Post: April 19th 2010, 12:53 PM
  5. Mathematical Modelling
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: April 16th 2009, 10:39 PM

Search Tags


/mathhelpforum @mathhelpforum