# differential calculus

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a pc of wire 10feet long is cut into 2 pcs. one piece is bent into the shape of a circle and the other into the shape of a square. how should the wire be cut so that the combined area is as large as possible?

• differential calculus -

Intuitively, 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
=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 d2A(x)/dx².

If d2A(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 -

This 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.