= 2xr.
But, you know that 400 = x + πr, or r = (400 - x) / π. Substitute this in:
Rectangular area
= 2xr.
= 2x {(400 - x) / π}.
= (2 / π) x (400 - x).
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1(c)
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Let's call the rectangle area R, so R = (2 / π) x (400 - x). Then, differentiate by the product rule:
dR
= d[(2/π) x (400 - x)].
= (2/π) d[x (400 - x)].
= (2/π) {x d[400 - x] + d[x] (400 - x)}.
= (2/π) {(x) (-1) + (1) (400 - x)}.
= (2/π) {-x + 400 - x}.
= (2/π) (400 - 2x).
= (4/π) (200 - x).
At a stationary point, dR = 0, and so:
dR = 0 =>
(4/π) (200 - x) = 0.
=> (200 - x) = 0
=> x = 200.
And using the relation between x and r:
400 = x + πr.
400 = 200 + πr.
200 = πr.
200/π = r.
To verify, differentiate dR, and if it is negative, then you have a maximum.
d[dR]
= d[(4/π) (200 - x)].
= (4/π) d[(200 - x)].
= (4/π) (-1).
= -4/π.
< 0.
Hence you have a maximum in the rectangle area at x = 200, r = 200/π.