Cone from a Circular Sheet
A metalworker has a circular sheet of radius 10 cm and wants to cut out a sector of angle θ (0<θ<2π), then join the two straight edges to form a cone. The slant height will be 10 cm, and the base radius r will be the arc length divided by 2π, so r = 10θ/(2π)=5θ/π. A draft calculation by an apprentice fixes θ=π, giving r=5 and height h=√(100−25)=5√3, and claims this must maximize volume because 'r and h are balanced.' The workshop needs the cone of largest volume, but the sheet can also be left partly unused; the sector angle may be any value in the open interval. The supervisor asks for a derivative-based check and also wants to know whether the endpoint cases θ→0 or θ→2π matter, since the forming process cannot handle a sector wider than 300° (θ≤5π/3). Determine the optimal feasible sector angle and compare it with the apprentice's choice.
1. Write the cone volume V as a function of θ, find the critical angle in the feasible interval 0<θ≤5π/3, and decide whether the apprentice's θ=π choice is optimal. Justify that the critical point is a maximum, not an endpoint.
Read the worked answer
Let x=θ/π, so 0<x≤5/3. Then r=5x and h=5√(4−x^2). Hence
V=(1/3)π(25x^2)(5√(4−x^2))=(125π/3)x^2√(4−x^2).
Set f(x)=x^2√(4−x^2). Differentiating,
f'(x)=x(8−3x^2)/√(4−x^2).
The only interior zero in 0<x≤5/3 is x=√(8/3)=2√6/3≈1.633, which is less than 5/3. Since f'>0 before this value and f'<0 after it, this critical point gives the maximum. Thus θ=π√(8/3)≈5.13 rad≈294°. The apprentice's θ=π gives x=1 and is not optimal. At x=5/3 the function is already decreasing, and as x→0, V→0, so neither endpoint is the maximum.
2. The apprentice claims that the maximum must occur when the base radius equals the height, because increasing one reduces the other. Test this claim under the cone constraint r^2+h^2=100, and compare the volumes at r=h and at the true optimum.
Read the worked answer
The claim is false. With r^2+h^2=100, write
V=(π/3)r^2√(100−r^2), 0<r<10.
Differentiate ln V:
2/r − r/(100−r^2)=0 ⇒ 2(100−r^2)=r^2 ⇒ r^2=200/3.
Thus r=10√6/3≈8.16 cm and h=10/√3≈5.77 cm, so r≠h at the optimum. At r=h=10/√2, V=(π/3)(50)(10/√2)≈370 cm^3. At the true optimum, V=(π/3)(200/3)(10/√3)≈403 cm^3. The true optimum is larger. The feasible sector limit r≤25/3≈8.33 still contains this r, so the boundary does not change the conclusion. A trade-off between variables does not imply equality at the maximum.