Putting differentiation into practice - A Level difficulty question

Please see attached diagram

An indoor running track consists of a rectangle with a semicircle of radius r at each end, as shown in the diagram. Given that the perimeter is to be 300 metres,

(a) Deduce the formula for the area A of the rectangular portion of the running track in terms of r.
(b) Find the dimensions of the running track that maximises the area of the rectangular portion and state the area enclosed within the entire perimeter of the running track.
(c) Plot a suitably labelled graph and interpret it to show how it verifies your findings.

any amount of help is very much appreciated, thanks.

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to be honest, i understand part (a), its just that i dont understand what is meant by the other 2 parts. i find calculus ok, but i find it difficult to put it into practice as above. thanks

btw i got A=300r - 2πr² for part (a) (in terms of r only)

q1.BMP161.7KB

Reply 1

arktos

a)

300 = 2x +2piR
150 - piR = x

area = 2Rx = 300R - 2piR^2

b)

dy/dR = 300 - 4piR = 0

R = 300/4pi = 23.87....

total area = piR^2 + x^2
total area = pi (569.93) + x^2

x^2 = (150- piR)^2 = 5625

total area = 7415.48 sq units.

sorry , area = piR^2 + 2Rx
total area = pi (569.93) + 2Rx
2Rx = 2 (23.87) (150-pi23.87) = 3580.986
total area = 3580.986 + 1790.007 = 5371 sq units.

Reply 2

Reloaded

Thanks.

What does it mean when it says "Find the dimensions of the running track that maximises the area of the rectangular portion"? Does that mean just set the dy/dr equal to zero and work out x and r?

also, how would i plot a graph and interpet it?

Anyone?

thanks

Reply 3

arktos

I wasn't sure of that one, I guess that you draw a diagram and put in the length of x and the circumference of the semi-circles?

for c) I would prob do a graph of y = 300x - pix²

with y as area and x as R

sorry I can't be of any concise help.

Reply 4

john!!

"Does that mean just set the dy/dr equal to zero and work out x and r?"

yes, or dy/dx, as the constraint on the perimeter allows you to eliminate one of these variables and find the area as a single variable function, so you might have the function y=A(r), and by setting A'(r) = 0 you maximise the area. Sometimes you need to show that A''(r)<0 to prove the point is a maximum and not a minimum or poi.

drawing the graph should be simple, except the questions seems a bit stupid bcause of its ambiguity, just sketch y=A(r).