sector of circle

Question

A sector with central angle θ is cut from a circle of radius R = 24 inches , and the edges of the sector are brought together to form a cone. Find the magnitude of θ such that the volume of the cone is a maximum.

Answer 1

This is one of the nicer optimization problems in Calculus 1, because nearly all the work is in translating the folding step into a relationship between the sector and the cone. Let me do that part slowly, since it is where people get stuck.

What the folding actually does. You cut a sector of radius R = 24 with central angle θ, then bring the two straight edges together. Two things carry over:

1. The straight edges of the sector become the slant height of the cone, so the slant height is 24.

2. The curved arc of the sector becomes the circular rim of the cone's base, so the arc length equals the base circumference.

That second fact is the key to the whole problem. The arc length is Rθ = 24θ and the base circumference is 2πr, so

24θ = 2πr,  which gives  r = 12θ/π

Get the height. The base radius, the height and the slant height form a right triangle with the slant height as the hypotenuse, so

h = √(24² − r²) = √(576 − r²)

Write the volume.

V = (1/3)πr²h = (1/3)πr²√(576 − r²)

It is much easier to optimize in terms of r and convert back to θ at the very end. That is legitimate because r and θ are related by a positive constant multiple, so whatever r maximizes the volume corresponds to exactly one θ.

A trick that saves you the square root. V is maximized exactly where V² is maximized, since V is positive and squaring is increasing on positive numbers. So maximize

V² = (π²/9)r⁴(576 − r²)

Drop the constant out front and differentiate g(r) = 576r⁴ − r⁶:

g'(r) = 2304r³ − 6r⁵ = 6r³(384 − r²)

Setting that equal to zero gives r² = 384, so r = √384 = 8√6 ≈ 19.60. Notice that 384 is exactly two thirds of 576, so r² = (2/3)R². That relationship is worth memorizing, because it is the answer to every problem of this shape no matter what the radius is.

Convert back to the angle. From 24θ = 2πr,

θ = 2πr/24 = πr/12 = π(8√6)/12 = 2π√6/3

θ = 2π√6/3 ≈ 5.130 radians ≈ 293.94 degrees

Does that answer make sense? It says keep a little more than 293 degrees of the circle and cut away a wedge of only about 66 degrees. That should feel right once you picture the two extremes. A very thin sector rolls up into a tall skinny cone with almost no base, and a nearly complete circle rolls up into a wide flat cone with almost no height. Both have tiny volume, so the best angle lives between them, and it turns out to sit much closer to the full circle than most people guess.

The cone itself, if you need it. r = 8√6 ≈ 19.60 in, h = √(576 − 384) = √192 = 8√3 ≈ 13.86 in, and V = (1/3)π(384)(8√3) = 1024√3 π ≈ 5572 cubic inches.