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GT
07-18-2006, 06:19 PM
hi this is a special question my teacher gave me see if you can help

a rectangular field is to be enclosed by a fence and then divided into 2 smaller plots by a fence parallel to one of the sides. find the dimensions of the field of greatest possible area if 1800 m of fence is available. state the total area

if you figure it out i will be very thankful

TchrWill
07-18-2006, 06:52 PM
hi this is a special question my teacher gave me see if you can help

a rectangular field is to be enclosed by a fence and then divided into 2 smaller plots by a fence parallel to one of the sides. find the dimensions of the field of greatest possible area if 1800 m of fence is available. state the total area

if you figure it out i will be very thankful

Let x = the width and y = the length

Then, A = x(1600 - 3x)/2 = (800x - 3x^2)/2

Take the first derivitive and solve for x and y.

Denis
07-19-2006, 01:00 AM
Armed with the knowledge(!) that a square gives max area, we can jump to the
conclusion that a square 360 by 360 = 129600 sq. ft. would be max area
(360 being also the inside fence length), but that would be wrong!

-------------w+x-----------
| | |
| | |
w w w
| | |
| | |
-------------w+x-----------

w = width; length => width, so length = w + x
To minimize inside fence, make inside fence = width (don't matter where it is)

w + w + w + w+x + w+x = 1800
simplify: x = (1800 - 5w) / 2 [1]

A = w(w + x) ; substitute [1]:
A = w(w + (1800 - 5w) / 2)
simplify: A = (1800w - 3w^2) / 2

Set to zero; take 1st derivative:
1800w - 3w^2 = 0
1800 - 6w = 0
w = 300 ; so, using [1], x = 150

A = 300 * 450 = 135000 sq. ft. max
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