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## Homework Statement

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?

## Homework Equations

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

## 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 :-)