Four feet of wire is to be used to form a square and a circle
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Four feet of wire is to be used to form a square and a circle

[From: ] [author: ] [Date: 11-06-27] [Hit: ]
Thus, we see that the minimum area is attained when x, the side length of the square, is 4/(4+(pi)), meaning the total amount given to the square is 16/(4+(pi)).Similarly,......
=(1/(pi))*( (4+(pi)) (16/(16+8(pi)+(pi)^2))-32/(4+(pi))+4)
=(1/(pi))*( (64+16(pi) )/(16+8(pi)+(pi)^2))-32/(4+(pi))+4)
This number is about .560.

Thus, we see that the minimum area is attained when x, the side length of the square, is 4/(4+(pi)), meaning the total amount given to the square is 16/(4+(pi)). Similarly, we see that the maximum area is attained when x is 0, giving all the length of the wire to the circle. (This mkes sense since a circle is the geometric shape with the greatest area/perimeter ratio.) So, the answers to the two parts above are:
(a) Give all 4 feet of wire to the circle and none to the square.
(b) Give 16/(4+(pi)) feet of wire to the square and the rest to the circle.
Hope that helped.

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You've specified the length of the wire.
What you seem to be asking is what form encloses the most area for a given perimeter.
It's the circle.
If by 'square', you mean 'rectangle', the square, (closest in configuration to the circle),
will enclose the most area for the 'rectangle' class of figures.

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Square of side s
circle of radius r

Objective: Area = s * s + pi * r * r
Constraint: Perimeter = 4 * s + 2 * pi * r = 4
=> s = 1 - pi * r / 2
Susbtitue
A = (1 - pi * r / 2) ^ 2 + pi * r ^ 2
= 1 + pi ^ 2 / 4 * r^2 - pi * r + pi * r ^ 2
= (pi + pi^2 / 4) * r ^ 2 - pi * r + 1
Take the derivate of A and set it to 0
A' = 2 * (pi + pi^2/4) r - pi = 0
So r = pi / (2 * (pi + pi^2 / 4))
And s = 1 - pi * r / 2
I am sure you can simplify it your self.

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Well... Obviously all four feet would be used in both.
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