View Full Version : word problem (HELP!!) 'a field is to be enclosed by a fence'

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, 05:52 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, 12:00 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

Powered by vBulletin® Version 4.2.2 Copyright © 2014 vBulletin Solutions, Inc. All rights reserved.