Page 2 of 3 FirstFirst 123 LastLast
Results 16 to 30 of 32

Math Help - Two problems

  1. #16
    Newbie
    Joined
    Mar 2010
    Posts
    18
    Quote Originally Posted by icemanfan View Post
    That's a start. When you multiply it out, you will find that the function is a parabola. Where does the parabola achieve its maximum value?
    by changing -100x^2-6000x+488000to f(x) = a(x-h)+k form ?
    Follow Math Help Forum on Facebook and Google+

  2. #17
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by myaa02 View Post
    by changing -100x^2-6000x+488000to f(x) = a(x-h)+k form ?
    Yes, you want to change the equation to that form. Be careful, though. You made a mistake in multiplying it out. The correct equation is f(x) = -100x^2 + 2000x + 480000.
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Newbie
    Joined
    Mar 2010
    Posts
    18
    Quote Originally Posted by icemanfan View Post
    Yes, you want to change the equation to that form. Be careful, though. You made a mistake in multiplying it out. The correct equation is f(x) = -100x^2 + 2000x + 480000.
    how did you get 2000x if you multiply 100x with 80 ?

    nvm i just got it lol
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Newbie
    Joined
    Mar 2010
    Posts
    18
    what do you do after -100(x^2-20x+100)+480000-100 ?
    Follow Math Help Forum on Facebook and Google+

  5. #20
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by myaa02 View Post
    what do you do after -100(x^2-20x+100)+480000-100 ?
    You should get f(x) = -100(x^2 - 20x + 100) + 480000 + 10000 = -100(x - 10)^2 + 490000.

    Then, observe that the parabola achieves its maximum value at the vertex. What is the vertex of this parabola?
    Follow Math Help Forum on Facebook and Google+

  6. #21
    Newbie
    Joined
    Mar 2010
    Posts
    18
    Quote Originally Posted by icemanfan View Post
    You should get f(x) = -100(x^2 - 20x + 100) + 480000 + 10000 = -100(x - 10)^2 + 490000.

    Then, observe that the parabola achieves its maximum value at the vertex. What is the vertex of this parabola?
    vertex is -10
    Follow Math Help Forum on Facebook and Google+

  7. #22
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by myaa02 View Post
    vertex is -10
    Incorrect. Remember, the vertex is a point. Also, the x-value of the vertex is not -10.
    Follow Math Help Forum on Facebook and Google+

  8. #23
    Newbie
    Joined
    Mar 2010
    Posts
    18
    Quote Originally Posted by icemanfan View Post
    Incorrect. Remember, the vertex is a point. Also, the x-value of the vertex is not -10.
    i seriously don't know how to get the vertex.
    Follow Math Help Forum on Facebook and Google+

  9. #24
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by myaa02 View Post
    i seriously don't know how to get the vertex.
    If the equation of the parabola is f(x) = a(x - h)^2 + k,
    then the vertex of the parabola is (h, k).
    Follow Math Help Forum on Facebook and Google+

  10. #25
    Newbie
    Joined
    Mar 2010
    Posts
    18
    Quote Originally Posted by icemanfan View Post
    If the equation of the parabola is f(x) = a(x - h)^2 + k,
    then the vertex of the parabola is (h, k).
    so its (-10,490000)
    Follow Math Help Forum on Facebook and Google+

  11. #26
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by myaa02 View Post
    so its (-10,490000)
    The vertex is (10, 490000).
    Follow Math Help Forum on Facebook and Google+

  12. #27
    Newbie
    Joined
    Mar 2010
    Posts
    18
    Quote Originally Posted by icemanfan View Post
    The vertex is (10, 490000).
    now how am I supposed to find the " new markup " in order to maximize the revenue ?
    Follow Math Help Forum on Facebook and Google+

  13. #28
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by myaa02 View Post
    now how am I supposed to find the " new markup " in order to maximize the revenue ?
    The function is maximized when x = 10. The markup is 100x, so what should the markup be?
    Follow Math Help Forum on Facebook and Google+

  14. #29
    Newbie
    Joined
    Mar 2010
    Posts
    18
    Quote Originally Posted by icemanfan View Post
    The function is maximized when x = 10. The markup is 100x, so what should the markup be?
    So, the markup should be 1000 ?
    Follow Math Help Forum on Facebook and Google+

  15. #30
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by myaa02 View Post
    So, the markup should be 1000 ?
    Yes, that's correct.
    Follow Math Help Forum on Facebook and Google+

Page 2 of 3 FirstFirst 123 LastLast

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: October 3rd 2011, 05:35 AM
  2. binomial problems/normal problems
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 19th 2010, 11:46 PM
  3. binomial problems as normal problems
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 19th 2010, 11:41 PM
  4. Problems with integration word problems
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 25th 2010, 05:39 PM
  5. Help thease problems problems
    Posted in the Geometry Forum
    Replies: 2
    Last Post: April 1st 2008, 11:03 AM

Search Tags


/mathhelpforum @mathhelpforum