Question 17784
let the length be x and the width be y
 then 2(x+y)=280 meters [length of fencing]
and xy=4000 sq m
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y=4000/x
substituting in eqn 1 we get,
2(x+y)=280
x+y=140
x+(4000/x)=140
x^2+4000=140x
x^2-140x+4000=0
(x-100)(x-40)=0
x=100 or 40
<br>
then y=4000/x
thus y=4000/100 or 4000/40
y=40 or 100
<br>
Hence possible dimensions are:
(length,width) in metres = (100,40) or (40,100)
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Hope this helps,
Prabhat