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November 24, 2014

November 24, 2014

Posted by **steph** on Thursday, November 10, 2011 at 10:20pm.

a) what is the length of the remaining piece of fencing in terms of x?

b) determine a function A that represents the total area of the enclosed region.

c) give any restrictions on x

d) what dimensions of the total enclosed region would give an area of 22,500 feet squared?

e) what is the maximum area that can be enclosed?

- pre-calc -
**Steve**, Friday, November 11, 2011 at 10:58am600 feet total fence

3 sides of x feet

remaining side: 600 - 3x

a(x) = x(600-3x) = 3x(200-x)

naturally, 3x < 600, so x < 200 assuming an infinitesimally thin fence and poles of zero diameter. :-)

22500 = 3x(200-x)

-3x^2 + 600x - 22500 = 0

x = 50 or 150

max area achieved at x = 100

a(100) = 30,000

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