A school is marking out a 400 m running track on its field. They want to be able to use the infield area (the rectangle between the semi-circles) to run PE games and activities. Due to this, they wish to maximise the rectangular area enclosed by the track. The track must consist of two straight sections and two semi-circular sections, and must be 400 m. Three examples of possible field configurations are shown below:
Find the dimensions of the rectangle that maximise its area.
Use calculus methods to show that this is a maximum.
Circumference of a circle: `C = 2pir`
We want to maximise the area of the rectangle, so we need an expression for its area. Start with the formula for the area of a rectangle, work out what the length (or base, or `x`) and width (or height, or `y`) are in terms of things we know, and substitute in those values.
Area of rectangle `= A = xy`
Perimeter `= 400` m
`{:("Rectangle length (width)",=,x,=,"half of total horizontal straight sections"),(,,x,=,1/2 ("perimeter"-"circumference" )),(,,x,=,1/2 (400 - 2pir)), (,,x,=,200 - pir),("Rectangle width (height)",=,y,=,2r):}`
We now have expressions for the rectangle's dimensions. Substitute in and simplify.
`A = 2r(200 - pir)`
`A = 400r - 2pir^2`
And differentiate with respect to `r`. We are also asked to show (with calculus methods) that it is a maximum, and we can do that now.
`A^' = (dA) / (dr) = 400 - 4pir = 0` at stationary point
`A^('') = (d^2A) / (dr^2) = - 4pi < 0 =>` a maximum
Great – whatever we find will be a maximum regardless of the value of `r`. Start work on solving the first differential set equal to zero, for r.
`100 - pir = 0`
`pir = 100`
`r = 100/pi ≈ 31.83` m
At this point it's best to keep using the exact value, rather than write the fraction as a decimal. It's also not a bad idea to check the value of `r`, but because of how we got the rectangle width, the two terms using `r` cancel each other out.
`2pir + 2(200 - pir) = 2pir + 400 - 2pir = 400`
Anyway, don't forget to reread the question after getting a result; we are asked for the rectangle's dimensions (not its area).
The rectangle is `x = 200 - pir = 100` metres wide.
The rectangle is `y = 2r = 2 times 100/pi = 63.66` metres high.
Comment: This seems a pretty standard question, although the assessment schedule does not require the height of the rectangle to be calculated or stated, or even that it's equal to `2r`. Just calculating `r` and the rectangle width is fine. And yet it does the value of the maximum rectangle area (in brackets) – to 6 significant figures! – even though that is not requested by the question. Weird.
