Calculus homework!!! Please help me?

Question

A sector with central angle X is cut from a circle of radius 4 inches, and the edges of the sector are brought together to form a cone. Find the magnitude of X such that the volume of the cone is a maximum.

Please help me and show all your work. I have no clue how to do this. Thank you! =)

Answer 1

volume of the cone = f(r) = (1/3)pi r^2 sqrt(4^2-r^2), where sqrt(4^2-r^2) is the height of the cone.

Solve f'(r) = 0 for r,
2r sqrt(4^2-r^2) + r^2(-r)/sqrt(4^2-r^2) = 0
=> 2(16-r^2) - r^2 = 0
=> r = sqrt(32/3)

Since the base circumference (2pi r) of the cone equals to the arc length (4x), we have,
4x = 2pi r
x = (1/2)pi sqrt(32/3) = 1.63 pi
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Ideas: The slant height (4 in.) of the cone is a constant. Therefore, we can optimize the radius (r) of the cone to get maximum volume.

Answer 2

The circumference of the base of the cone will be 4X, with 4 being the lateral height of the cone.
r = 4X/(2π) = 2X/π, dr/dX = 2/π
h = √(16 - r^2)

Edit:

V = (1/3)πr^2√(16 - r^2)
dV/dr = (1/3)((- 2r)(1/2)πr^2/√(16 - r^2) + 2πr√(16 - r^2))
2πr√(1 - r^2) = πr^3/√(16 - r^2)
2(16 - r^2) = r^2
32 - 2r^2 = r^2
3r^2 = 32
r^2 = 32/3
2X/π = √(32/3)
X = (π/2)√(32/3) ≈ 3.265986 rad ≈ 187.1272°

Check:
Vmax = (1/3)π(32/3)√(16 - 32/3) ≈ 25.79626
V+ = (1/3)π(33/3)√(16 - 33/3) ≈ 25.75765
V- = (1/3)π(31/3)√(16 - 31/3) ≈ 25.75923

Answer 3

None of the homework help answers are 100% correct!!