Given a rectangle with sides W and H
The perimeter P=2W+2H
The area A=WH
If A is fixed how do I show the minimum P occurs when W=H (ie a square)?
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Timothy
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Timothy
I've got it:
A=WH
P=2W+2H
W=A/H
P=2(A/H)+2H
dP/dH=-2A/(H^2)+ 2
min at 0
0= -2A/(H^2)+2
H^2 = A
Therefore H=root A for min P
ie A square
How many marks is that worth in an essay about minimising building
costs?!
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Timothy
Substitute H=A/W into the first equation then solve the resulting
quadratic?
P=2W+2A/W -> PW=2W^2+2A -> 0=2W^2-PW+2A, then solve as appropriate.
Stu
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From the prompt of Stu Teasdale
I know some very good answers have been given already, but here's how I'd
have done it...
Rather than considering the relationship between the full perimeter, work
from (x=)P/4; which would of course be the length of the side of a square
for the given area, W and L then become x +/- d:
A = W . H = (x-d)(x+d) = x^2 - d^2
Now, can anyone remember which value of d gives the lowest possible value of
d^2? :)
Cheers
- Darren
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Darren Edmundson - Internet & 3G Technologist
MSN:grid...@hotmail.com
i
HTH
Andy Scheller
<can't be bothered with a sig>
Or even 54281905383476309103852764j?
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yours aye,
Tom
j? whassat? talk maths, not engineer!
star