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posted by  Novaworks on 11/2/2009 8:51:49 PM  |  status: Closed  |  Earned Karma: 1145

Find the dimensions...

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Calculus N/A N/A N/A 11/3/2009 at 2:00:00 PM
Question Details:
Find the dimensions of a rectangle with area 27000 m2 whose perimeter is as small as possible. (Give your answers in increasing order, to the nearest meter.)
1?  m
2? m
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Oracle
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posted by Altair on 11/2/2009 8:59:37 PM  |  status: Live
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Response Details:
let the side be x and y
the area A = x*y = 27000 m2
=>y = 27000m2/x
the parameter is: P = 2x + 2y = 2x + 54000 m2/x
P' = 2 - 54000/x2
let P'=0, we have: 2 - 54000/x2 =0
=>x = √(27000m2) = √(27000)  m ˜ 164.32 m
   y = 27000 m2/x = √(27000) m  ˜ 164.32 m
it is a square.

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posted by new_sylar on 11/3/2009 4:21:11 AM  |  status: Live
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Response Details:
let length be l and width be b

the lb=27000

h(x) =2l+2b to be maximum

consider lagrange k(l,b) =2l+2b-lb-27,000

for this to be maximum k(l,b) derivative must be zero. so which gives  l=b
so h(x) =4b =4*√27000 =657m

small trick is consider square for these type of problems
-------Ford :p
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