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Problem with differentiation to find maximum points

  1. Dec 2, 2011 #1
    1. The problem statement, all variables and given/known data

    A sports stadium is lit by four floodlights standing at the four corners of a rectangle which contains the rectangular pitch placed symmetrically inside it. The length of the rectangle is 160 metres and the width is 64 metres. This question is concerned with finding the common optimal height for the floodlights giving 'best' illumination of the pitch.

    A coordinate system is set up with the origin at the centre of the pitch. The x axis points along the pitch and the y axis points across the pitch.

    The luminance I produced at a point by a single light of power q positioned at a height h above the point (x0, y0) on the pitch is given by:

    I = (q*h)/((h^2 + (x-x0)^2 + (y-y0)^2 )^3/2)

    The luminance at any point on the pitch is given by the sum of the luminances at that point from each light. The power for each light is 1,730,600 units

    i) Find the value of h for which the luminance at the centre of the pitch takes its maximum value.

    ii) With the floodlights constructed at this optimal value of h, give an exact value of the coordinate of the darkest point.

    iii) Give an exact value of the y coordinate of the darkest point.

    iv) It is decided to adjust the common height of the floodlights in order to give maximum luminance at the darkest points. At what height should the lights be placed?

    v) With the floodlights set at this revised height, what is the luminance at the darkest point?

    2. Relevant equations

    I = (q*h)/((h^2 + (x-x0)^2 + (y-y0)^2 )^3/2)

    3. The attempt at a solution

    i) So for the first part i differentiated I with respect to h to find a value for h that maximised I at the centre of the pitch. i set q=1730600, y=0, x=0, y0=64/2, x0=160/2. The lights are at the corners of the pitch so the coordinate system means that the positions of the lights are (80, 32), (80,-32), (-80, 32), (-80, -32)
    i then said dI/dh=0 and solved that for h and i got a value of 60.926

    ii & iii) the darkest points will be at (0, 32) and (0, -32)

    then for part iv im not really sure what to do... i guess you would differentiate I again with respect to h but this time say that y=32, x=0?? but since its no longer in the centre wont the values of x0 and y0 be different for each of the lights?

    any help would be appreciated :-)
  2. jcsd
  3. Dec 8, 2011 #2
    anyone?? :-(
  4. Dec 9, 2011 #3

    Ray Vickson

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    By looking at 3d plots done in Maple, I find that for small h the darkest spots are in the center of the short side, but for larger h they are at the corners. (I don't know whether you regard looking at plots as a "proof" or not, but you can also verify the solution by using optimality conditions in the presence of constraints---where not all derivatives will equal zero!) Anyway, you can find expressions for the light levels at the centers of the short sides and at the corners, as functions of h, find the h-interval in which one is larger than the other and so in that way find the largest value of the smaller of the two.

  5. Dec 9, 2011 #4
    Hi Ray :-)
    what is Maple? i have Mathcad v15 but i dont really understand how to use the 3d plots feature. is Maple any better?
    So if i get expressions for the luminance at the both points, the centre of the side and at the corners, what do you then mean by 'find the h-interval'? and how would i start to do this?
  6. Dec 9, 2011 #5

    Ray Vickson

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    Maple is an alternative to Mathematica, with about the same functionality. I don't have access to Mathcad, so I can't help you with plotting.

    Anyway, my response was not correct; it was based on looking at a portion of the plot instead of the whole plot. Here is the revised situation: for ANY h > 0 the minimum is in the center of the LONG sides. It is easy enough to obtain the illumination at the long sides' centers, and then adjust h to maximize it, at least numerically. Now there is no need to consider "intervals"; there is just one expression now, and all of {h > 0} is the interval.

  7. Dec 9, 2011 #6
    yeah i get that the darkest points as being in the centre of the long sides. so if i differentiate the original equation with respect to h and then set x=0, y=32, x0=80, y0=32 and q=1730600 and solve the equation equal to zero then i will get a value for h that maximises the luminance at these points right?
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