Posted by **lisa** on Saturday, March 2, 2013 at 9:41pm.

Suppose you have 400 meters of fencing and you wish to enclose a rectangular space with two parallel partitions: What is the maximum total area of the enclosed space, in square meters?

- calculus -
**Reiny**, Sunday, March 3, 2013 at 12:01am
If the original rectangle is x m wide and y m long

then the partitions are x m as well

so we have

2y + 4x = 400

y + 2x = 200

y = 200 - 2x

Area = xy

A = x(200-2x) = 200x - 2x^2

dA/dx = 200 - 4x = 0 for a max of A

200 = 4x

x = 50

then y = 200 - 100 = 100

if x = 50 and y = 100, we get a maximum area

which is xy or 50(100) = 5000 m^2

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