This pops up in many forms. If the length of the wire is z, the side of the triangle is s, and the radius of the circle is r, then we have
3s + 2πr = z
so, s = (z-2πr)/3
The total area is
A = √3/4 s^2 + πr^2
= √3/36 (z-2πr)^2 + πr^2
You want A to be minumum, so dA/dr = 0
dA/dr = √3/36 * 2(z-2πr)(-2) + 2πr
= 2πr - π/√27 (z-2πr)
= (1+π/√27)(2πr) - 2πz/√27
dA/dr = 0 when r = z/(π+3√3) ≈ 8.51
The area of the circle is
πr^2 = π*(71/(π+3√3))^2 ≈ 227.81
double-check my math there.
You have a wire that is 71 cm long. You wish to cut it into two pieces. One piece will
be bent into the shape of a right triangle with legs of equal length. The other piece
will be bent into the shape of a circle. Let A represent the total area enclosed by the
triangle and the circle. What is the circumference of the circle when A is a minimum?
1 answer
2 answers