Question

Need help please!!!
A piece of wire 4 m long is cut into two pieces. One piece is bent into the shape of a circle of radius r and the other is bent into a square of side s. How should the wire be cut so that the total area enclosed is:

a) a maximum? r= and s= .

b) a minimum? r= and s=

Answer 1

The total area enclosed by the shapes is \(A=\pi r^2+s^2\), and is constrained by the fact that the total length of the wire is \(4\). This means the circumference of the circular piece and the perimeter of the square piece must add up to \(4\), so the constraint is \(2\pi r+4s=4\).

Out of curiosity, do you know about Lagrange multipliers or is this still single variable calculus?

Answer 2

single variable calculus

Answer 3

I solved for s in the 2pir+4s=4 equation and then put in the area equation. I then took the derivative and solved for r and got 2/5, but it was wrong

Answer 4

Okay, so you have
\[2\pi r+4s=4~~\implies~~s=\frac{2-\pi r}{2}~~\implies~~A=1-\pi r+\frac{4\pi+\pi^2}{4}r^2\]which gives
\[\frac{dA}{dr}=-\pi+\frac{4\pi+\pi^2}{2}r=0~~\implies~~r=\frac{2}{4+\pi}\]

Answer 5

yes

Answer 6

Glad you agree?