Friday

July 25, 2014

July 25, 2014

Posted by **jr** on Wednesday, July 10, 2013 at 4:37am.

- differential calculus -
**MathMate**, Wednesday, July 10, 2013 at 6:11amIntuitively, the circle has a bigger area for the same circumference, so we try to make a bigger circle.

Also, since the area increases as the square of the circumference, we want to make 1 single shape as opposed to two.

So intuitive answer is to make the smallest square possible (zero if allowed).

Let's check our intuition.

Let x=length to make the circle

Radius of circle, r

=x/(2π)

Area of circle

=πr²

=x²/(4π)

For the square,

perimeter=10-x

side length, s

=(10-x)/4

Area of square

=((10-x)/4)²

Total area, A(x)

=x^2/(4*π)+((10-x)/4)²

dA(x)/dx=x/(2π)-(10-x)/8

For maximum, equate dA(x)/dx to zero, and solve for x.(=4.399)

Check if the answer is a maximum or minimum by graphing, or calculate d^{2}A(x)/dx².

If d^{2}A(x)/dx².>0, it is a minimum, and thus has no use for us.

In that case, we calculate A(0), A(4.399) and A(10) to find the areas at the critical points and choose the largest one.

- differential calculus -
**Reiny**, Wednesday, July 10, 2013 at 8:57amThis is a very common question in introductory Calculus, and each time I have seen it, it asked for a MINIMUM area , and as MathMate noted, we would make the circle as large as possible.

Taking it with the usual minimum area request,

let the radius of the circle be r

let the side of the square be x

then 2πr + 4x = 10

x = (5 - πr)/2

A = πr^2 + x^2 = πr^2 + (1/4) (25 - 10πr + π^2 r^2)

dA/dr = 2πr + (1/4)(-10 +2π^2 r) = 0 for a min of A

dividing by π and solving for r,

r = 5/(4+π) = appr .7001

x = 1.4002

since 4x is needed for the square, the wire should be cut

appr 5.6 cm for the square and 4.4 cm for the circle.

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