Find the magnitude of 𝜃 such that the volume of the cone is a maximum.

Question

A sector with central angle 𝜃 is cut from a circle of radius R = 18 inches (see figure), 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

v = πr²h/3; 2πr = 18θ; r = 9θ/π; h = √(18² - 81θ²/π²) = 9√(4 - θ²/π²).

v = π·81θ²/π²·3√(4 - θ²/π²) = 243θ²√(4 - θ²/π²)/π;

v' = 243/π[2θ√(4 - θ²/π²) - (θ³/π²)/√(4 - θ²/π²)] = 0; θ = 0 or 2(4 - θ²/π²) - θ²/π² = 0; 8 - 3θ²/π² = 0;

θ = ±2√6π/3. Let take positive value for θ = 2π√6/3. To prove that this value of θ gives max you can use

fact that v(0) = 0 is minimum

Answer 2

Let R=18 inches

Let Φ=2π-θ

Where θ is the angle cut out of the circle and Φ will be the Circumference of the top of the cone,

C=RΦ

Let r = the radius of the top of the cone, r = C/(2π) = RΦ/(2π)

Volume of a cone is, V= πr²h

Need to find h. R is the hypotenuse of a right triangle, h²=R²-r² = R²-R²Φ²/(4π²)

V(Φ)= π(R²/(12π)) Φ²R√(1-Φ²/(4π²))

To minimize, take the derivative and then set that to zero and solve for Φ.

V'(Φ) = (R³/(12π)) (1/√(1-Φ²/(4π²)))(2Φ-3Φ³/(4π²))

2Φ-3Φ³/(4π²) =0 Notice that R doesn't matter

Φ²=8π²/3

Φ=π√(8/3)

θ = 2π-Φ = π(2-√(8/3)) ≈ 0.367π ≈ 66.06°

Checking if maximum using first derivative test

V(360-66.06) ≈ 7384.9 <-- max

V(360-65) ≈7384.3

V(360-67)≈7384.4

Answer 3

angle that maximizes volume = 90 degrees

(this answer is "under construction," with expected changes,soon)

it might help to draw a diagram, 3D, of a cone, or even cut out a circle

and then cut out an angle from it, about 90 degrees, 1/4 of the circle, to better visualize the problem

even scotch tape what's left into a cone

3 or so solutions below, all probably wrong, except number 2

volume =(1/3)hpir^2/3, r=18sin(T/2), h = 18cos(T/2)

v(T) = (1/3)(18cos(T/2)(18sin(T/2))^2

= (1/3)(18^2)cosT/2(sin^2(T/2))

= 108cos(T/2)sin(T/2)sin(T/2)

= 108(1/2)sinT(sin(T/2)

= 54sinTsin(T/2)

v'(T) = 54(sinT(cos(T/2)(1/2) +sin(T/2)cosT)

= 27(sinTcos(T/2) + cosTsin(T/2))

= 27(sin(T+T/2))

sin(T+T/2) = 0

3T/2 = 0 or 180

T = 2/3 x 180 = 120 degrees = optimal angle

or,

another solution

v= (1/3)(18^2)sin^2(T/2)cos(T/2)

= (18^2)/3][sin(T/2)(sin(T/2)cos(T/2)

= 6(18)(1/2)sinTcos(T/2)

= 54(sin(T)cos(T/2)

=54((sqr(1-cos^2(T)/2))cos(T/2)

v'=54[(1-cos^2(T))/2)(1/2)(-sin(T/2)) + cos(T/2)(1/2)(1/2)(-sin(T/2))]=0

(-sin(T/2)[sqr(1-cos^2(T/2)) + cos(T/2)] = 0

set factors =0

1st factor gives a minimum, not max

2nd factor

leads to

(sqr(1-cos^2(T/2) + cos(T/2) =0

sqr(1-cos^2(T/2) = - cos(T/2)

square both sides

1-cos^2(T/2) = cos^2(T/2)

let x = cos(T/2)

1- x^2 = x^2

2x^2 = 1

x^2 = 1/2

x = sqr(1/2) = sqr2)/2

cos^(T/2) = (sqr2)/2

T/2 = 45 or 315 degrees

T = 90 degrees = optimal angle to maximize volume

there's an error in, at least, one of the above answers

2nd answer, 90 degrees, looks more reasonable

with T=90

v(90) = 108(sqr2)/2 = 54sqr2 = about 76.37 ft^3 = max volume

v(120) = 108cos120sin^2(120) <0, leading to a negative volume which is not a maximum volume

as cos120 =-1/2, while sin^2(120) = 3/4

108(-1/2)(3/4) = -40.5 ft^3

another method is calculate the circumference of the top of the cone

by calculating the arc length after removing the sector's arc with angle T

the circle's circumference = pi(2)(18)= 36pi

the sector's arc = 18T

36pi-18Ti = (18)(2pi-T)

then calculate the radius of the cone top

C=2r = 18(2pi-T)

r = 18pi-9T

Area of the cone top = pi(r^2) = pi(18pi-9T)^2

Volume = (1/3)height x A

= (h/3)A = h(18pi^2 -9piT)

h= 18cos(T/2)(18pi^2 -9piT)

V = (1/3)cos(T/2)(18pi^2 -9piT)(pi(18pi-9T)^2

= cos(T/2)(6pi^2 - 3piT)(pi)(18pi-9T)^2

=3pi(cos(T/2)(2pi - T)(pi)(18pi-9T)^2

= 3pi^2(cos(T/2)( ...

between 2 other tutor's different answers

and a few more above, you have a buffet of possible answers

best answer is carefully, very carefully, doing the arithmetic and algebra

along with trig and calc

v = 18cos(T/2)54pi^2 -18(3)piT

= 972pi^2(cos(T/2) - 54piT

v' = 972pi^2(1/2)(-sin(T/2) -54pi = 0

-9sin(T/2) =1

sin(T/2) = -1/9

T/2 = 6.3793

T = about 12.7586 degrees

v(12.8) = about 54(cos6.4)sin^2(6.4)

= about 2/3 ft.^3

the closer the angle T approaches zero, the volume approaches zero

same with T approaching 360, the volume approaches zero

optimal volume maximizing T is between 0 and 360

likely not anywhere near 0 or 360

more likely between about 90 and 270, 90<T<270

find the error(s) in the above and you get 2 extra credit points