A paper cone is to be formed by starting with a disk of radius 9cm, cutting out a circular sector, and gluing the new edges together. The size of the circular sector is chosen to maximize the volume of the resulting cone. How tall is the cone?
To find the height of the cone, we first need to determine the dimensions of the circular sector that will result in the maximum volume.
Let's assume the angle of the circular sector is θ (in radians).
The circumference of the resulting cone will be the same as the circumference of the original circle, which is 2π(9) = 18π cm.
Since the circumference of a circle is given by 2πr, the arc length of the circular sector can be calculated using the formula:
arc length = θr
So, the arc length of the circular sector is θ(9) = 9θ cm.
The slant height of the cone can be calculated using the Pythagorean theorem. The slant height represents the distance from the apex of the cone to the edge of the circular sector.
We can form a right triangle by drawing a line from the center of the original circle to the midpoint of the arc. This line is also the height (h) of the resulting cone.
The slant height (s) and the radius (r) of the original circle form the legs of the right triangle. So, according to the Pythagorean theorem, we have:
s^2 = r^2 + h^2
Since r = 9 cm and s = 9θ cm, we can substitute these values into the equation:
(9θ)^2 = 9^2 + h^2
81θ^2 = 81 + h^2
h^2 = 81θ^2 - 81
h^2 = 81(θ^2 - 1)
Now, let's determine the volume of the cone. The volume (V) of a cone is given by the formula:
V = (1/3)πr^2h
Substituting the value of r = 9 cm, the volume equation becomes:
V = (1/3)π(9^2)h
V = 27πh
To maximize the volume, we need to maximize the value of h. Since h^2 = 81(θ^2 - 1), maximizing h is equivalent to maximizing θ^2 - 1.
To find the maximum value of θ^2 - 1, we need to find the critical points of this function. Taking the first derivative with respect to θ and setting it equal to zero will help us find these points.
d(θ^2 - 1)/dθ = 2θ = 0
So, θ = 0 is the only critical point.
Now, let's find the second derivative of θ^2 - 1 to determine the nature of this critical point.
d^2(θ^2 - 1)/dθ^2 = 2
Since the second derivative is positive, this critical point represents a local minimum. However, we are interested in the maximum value, so we can ignore this critical point. This means that the value of θ that maximizes the volume does not have a critical point within the domain of interest.
Therefore, we need to determine the maximum value of h by finding the maximum value of θ^2 - 1 within the allowed domain.
Since θ can only range from 0 to 2π, we evaluate θ^2 - 1 at the endpoints of this interval:
For θ = 0: h^2 = 81(θ^2 - 1) = 81(0^2 - 1) = -81
For θ = 2π: h^2 = 81(θ^2 - 1) = 81((2π)^2 - 1) = 81(4π^2 - 1)
Calculating these values, we find that h^2 is negative at the lower endpoint and positive at the upper endpoint. Hence, the maximum value of h^2 occurs at the upper endpoint, which means the maximum volume is obtained when θ = 2π.
Substituting this value of θ into the equation h^2 = 81(θ^2 - 1), we get:
h^2 = 81((2π)^2 - 1) = 81(4π^2 - 1)
h^2 = 81(16π^2 - 1)
h = √(81(16π^2 - 1))
Finally, we can calculate the value of h:
h ≈ √(81(16π^2 - 1)) ≈ 36.484 cm
Therefore, the height of the cone is approximately 36.484 cm.