Results 1 to 8 of 8

Math Help - Fitting data to a model

  1. #1
    Newbie
    Joined
    Mar 2009
    From
    USA
    Posts
    13

    Fitting data to a model

    given the following set of data where x represents the number of items sold and y(x) represents the profit when x items are sold
    X # of items Y(profit)
    0 -1350
    1 -800
    2 -350
    3 0
    4 250
    5 400
    6 450
    7 400
    for letter A i already got the answer the quadratic formula which is y=-50x+600x+350
    B. create a specific algebraic model for profit as a function of the number of items sold.show your work and clearly define variables
    C. the break-even point is when profit is zero. what is/are the break even points?explain your answer
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2
    Hyvää on! Mistä häiriöstä?

    y = -50x^{2} + 600x - 1350

    Olkaa varovaisia!

    B: Isn't that what you did in A? Don't forget x \; \epsilon \; [0,7]

    C: Solve! 0 = -50x^{2} + 600x - 1350
    Follow Math Help Forum on Facebook and Google+

  3. #3
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by kenkarylle5 View Post
    given the following set of data where x represents the number of items sold and y(x) represents the profit when x items are sold
    X # of items Y(profit)
    0 -1350
    1 -800
    2 -350
    3 0
    4 250
    5 400
    6 450
    7 400
    for letter A i already got the answer the quadratic formula which is y=-50x+600x+350
    B. create a specific algebraic model for profit as a function of the number of items sold.show your work and clearly define variables
    C. the break-even point is when profit is zero. what is/are the break even points?explain your answer
    Aren't A and B one and the same?

    Did you mean -50x^2+600x-1350 = 0? Otherwise f(0) isn't correct nor is it a quadratic as you've written it.

    Cancel out a factor of 50 to make this simpler to solve and move everything to the other side (you don't have to do either of these but I prefer having positive a)

    50x^2-600x+1350 = 0 \rightarrow x^2 - 12x + 27 = 0

    Solve using your favourite method (I get x=3 and x=9 as solutions)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2009
    From
    USA
    Posts
    13
    i got B, but im not sure with C, am i going to base my answer on the chart for letter C...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2
    If you have a perfect model, which you do, the chart or the equation will do just fine. Since this is a quadratic model, there will be another zero for somewhere x > 7. You will have to decide if this is a valid result. Is the model defined for x > 7? Perhaps only 7 can be produced?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Mar 2009
    From
    USA
    Posts
    13
    so my break-even point when profit is zero is 7 then huh.?
    mann im sorry, im really stuck, but thanks for your helps guys appreciate it..
    Follow Math Help Forum on Facebook and Google+

  7. #7
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by kenkarylle5 View Post
    so my break-even point when profit is zero is 7 then huh.?
    mann im sorry, im really stuck, but thanks for your helps guys appreciate it..
    No, 7 may be your upper limit on the domain rather than being a 0 point. It means that you can disregard any solutions above 7.

    This won't apply if your domain (x value) can go higher than 7
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Mar 2009
    From
    USA
    Posts
    13
    ah okay... tnx, now i will work on 2 more word problems.tks tsk tsk
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Data fitting
    Posted in the Statistics Forum
    Replies: 2
    Last Post: April 28th 2010, 11:51 PM
  2. fitting polynomial functions into data
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 21st 2010, 01:37 PM
  3. Fitting a hyperbola to data
    Posted in the Math Topics Forum
    Replies: 6
    Last Post: February 9th 2010, 12:17 AM
  4. simple data fitting problem
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: October 7th 2009, 07:10 AM
  5. Replies: 3
    Last Post: April 15th 2009, 01:18 AM

Search Tags


/mathhelpforum @mathhelpforum