Results 1 to 4 of 4

Math Help - Applied Optimization Problem

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    31

    Applied Optimization Problem

    Let Q = (0,4) and R= (9,7) be given points in the plane. We want to find the point P=(x,0) on the x-axis such that the sum of distances PQ+PR is as small as possible.

    f(x) = ?

    [a,b] a = ? b = ?

    The minimal sum of distances is ?


    Can anyone show me how to work this, as I am utterly lost. All I can muster is maybe sqrt(-x^2 + 16) + sqrt((9-x)^2+49). I'm all fuggled.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    There's a little trick that will get the answer quickly. Imagine that the x-axis is a mirror, so that the point R is reflected to the point R'=(9,-7). The shortest distance from Q to R' is a straight line, and you can easily find the point P where that meets the x-axis. Then PQ+PR=PQ+PR', and that will give the solution to the problem.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by pseizure2000 View Post
    Let Q = (0,4) and R= (9,7) be given points in the plane. We want to find the point P=(x,0) on the x-axis such that the sum of distances PQ+PR is as small as possible.

    f(x) = ?

    [a,b] a = ? b = ?

    The minimal sum of distances is ?


    Can anyone show me how to work this, as I am utterly lost. All I can muster is maybe sqrt(-x^2 + 16) + sqrt((9-x)^2+49). I'm all fuggled.
    PQ = sqrt[(x-0)^2 +(0-4)^2] = sqrt[x^2 +16]
    PR = sqrt[(x-9)^2 +(0-7)^2] = sqrt[(x-9)^2 +49]

    Let D = PQ +PR
    So,
    D = sqrt[x^2 +16] +sqrt[(x-9)^2 +49] ---------------***

    For maximum or minimum D, dD/dx = 0

    dD/dx = [x /sqrt(x^2 +16)] +[(x-9) /sqrt((x-9)^2 +49)]

    Set that to zero,

    0 = [x /sqrt(x^2 +16)] +[(x-9) /sqrt((x-9)^2 +49)]

    Mutiply both sides by the product of the two denominators,

    0 = x*sqrt((x-9)^2 +49) +(x-9)sqrt(x^2 +16)

    -x*sqrt((x-9)^2 +49) = (x-9)sqrt(x^2 +16)

    Square both sides,

    (x^2)[(x-9)^2 +49] = [(x-9)^2](x^2 +16)

    (x^2)(x-9)^2 +49x^2 = [(x-9)^2](x^2 +16)

    49x^2 = [(x-9)^2](x^2 +16) -(x^2)(x-9)^2

    49x^2 = [(x-9)^2](x^2 +16 -x^2)

    49x^2 = 16(x-9)^2

    Take the square roots of both sides,

    7x = +,-4(x-9)

    7x = 4(x-9)
    7x = 4x -36
    7x -4x = -36
    x = -12 ---------------(i)

    7x = -4(x-9)
    7x = -4x +36
    7x +4x = 36
    x = 36/11 = 3.272727.... --------(ii)

    If x = -12,
    D = sqrt[x^2 +16] +sqrt[(x-9)^2 +49] ---------------***
    D1 = sqrt[(-12)^2 +16] +sqrt[(-12 -9)^2 +49]
    D1 = sqrt[160] +sqrt[490]
    D1 = 34.78 unit long.

    If x = 3.272727....,
    D = sqrt[x^2 +16] +sqrt[(x-9)^2 +49] ---------------***
    D2 = sqrt[(3.272727...)^2 +16] +sqrt[(3.272727.. -9)^2 +49]
    D2 = sqrt[26.71] +sqrt[81.80]
    D2 = 14.21 unit long.

    So, D1 is maximum, and D2 is minimum.

    Therefore, for (PQ +PR) to be as small as possible, the P must be at (36/11,0). ---answer.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2007
    Posts
    31
    wow this is great work and 100% correct. thanks, it was quite mind boggling to me.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Applied optimization in calculus
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 23rd 2009, 10:00 PM
  2. [SOLVED] applied optimization problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 3rd 2009, 05:40 PM
  3. Applied Optimization Problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 9th 2008, 09:34 PM
  4. Applied Optimization
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 22nd 2008, 01:26 PM
  5. Need Help: Applied Optimization
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 7th 2007, 09:20 PM

Search Tags


/mathhelpforum @mathhelpforum