r2 = h2 + (rθ/2π)2
h2 = r2  (rθ/2π)2
h = √ [r2  (r2θ2/4π2)]
Therefore, substituting the values for rbase and h, we can find the volume of the cone.
V = 1/3 x height x base area
V = 1/3 × √ [r2  (r2θ2/4π2)] × π (r2θ2/4π2)
2. By using the substitution x = θ/2π , express the volume as a function of x.
x = θ/2π
θ = x2π
V = 1/3 × √ [r2  (r2θ2/4π2)] × π (r2θ2/4π2)
V = 1/3 × √ [r2  (r2(x2π)2/4π2)] × π (r2(x2π)2/4π2)
= 1/3 × √ [r2  (r24x2π2/4π2)] × π (r24x2π2/4π2)
= 1/3 × √ [r2  (r2x2)] × π (r2x2)

Draw the graph of this function using the calculator. Hence find the values of x and θ for which this volume is a maximum. Give your answer for x to four decimal places.
Since x = θ / (2π), the maximum x value will be a constant no matter what the value of r is (x represents the amount per proportion of paper used for creating the cone, which is a constant ratio. The value of r will change the volume of the cone, but x will remain the same). Therefore we can replace the r in the function with any constant number. And in this project, 1 is used for replacing r since it makes the function easier to calculate. Thus:
Vcone = 1/3 × √ [r2  (r2x2)] × π (r2x2)
Vcone = 1/3 × √[12 – (x212] × π x212
Vcone = 1/3 × √(1 – x2) × π x2
In this function, 1/3 π is constant. The shape of the function will be x2 × √(1 – x2)
To find the maximum value of the cone, we can either:
Graph Vcone = x2 × √(1 – x2) and find the maximum value or
Graph the derivative of Vcone and find the zero/root of the function
Graph of Vcone (grey) and its derivative (orange)
The two maximum values are at x = ±0.8174. They both give a maximum cone volume of 0.3871.
To calculate the value of θ, for which the cone volume is a maximum:
x = θ / (2π) = ±0.8174
Thus, θ = 2πx = 2π × ±0.8165 = ±5.136 rad
When the circle is cut there are two sectors so it is possible to make two cones.
 Find the value or values of x for which the sum of the volumes of the two cones is a maximum.
To find the sum of the volumes of the two cones, we must find the volume of each individual cone. We will call them Cone A and Cone B.
From Question 1, we know the volume of Cone A:
Vcone (A) = 1/3 × √(r2 – (x2r2) × π x2r2
The angle of Cone B = 2π  θ
xB = θ / 2π
= 2π  θ / 2π
= 1  θ / 2π
= 1 – x
The volume of Cone B can be expressed as:
Vcone (B) = 1/3 × √[r2 – (1x)2 r2] × π (1x)2 r2
Finding the sum of the two cones:
V = Vcone (A) + Vcone (B)
= 1/3 × √(r2 – x2r2) × π x2r2 + 1/3 × √[r2 – (1x)2 r2] × π (1x)2 r2
Like Question 3, we can replace the constant r with 1 for easier calculation.
We get: V = 1/3 √(1 – x2) × π x2 + 1/3 × √[1 – (1x)2 ] × π (1x)2
Graphing this equation, we can find the maximum sum of the volumes of the two cones.
Graph of Vcone (A) (grey), Vcone (B) (red), and Vsum (blue).
The two maximum cone volumes we get are both 0.4566. The x values that give this maximum cone volume are 0.6760 and 0.3240.
Notice that these two x values are just complements of each other (they add up to 1). This means that x = 0.3240 and x – 1 = 0.6760. The maximum volume = 0.6760.
Similarly, this answer can also been found by finding the derivative of Vsum and equating it to zero. The zero/root of the function will be the maximum value.