# Volumes of Cones

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Introduction

Diana Herwono D 0861 006

IB Mathematics HL

Portfolio

(Type III)

## Volumes of Cones

By Diana Herwono

IB Candidate No: D 0861 006

May 2003

In this assignment, I will use WinPlot, a graphing display program.

1. Find an expression for the volume of the cone in terms of r and θ.

The formula for the volume of a cone is: V = 1/3 x height x base area

To find the base area, we must find the radius of the base.

The circumference of the base of the cone = the length of arc ABC

Therefore:

2π × rbase = rθ

rbase = rθ / (2π).

The area of the base can therefore be calculated:

Abase = π × rbase2 = π × ( rθ / (2π))2

Next, we must find the height of the cone, h.

Notice that for the cone, hrbase r is a right angled triangle, with r as the hypotenuse.

Therefore, using the Pythagorean Theorem, we can find h.

Middle

= 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

Conclusion

Finding the sum of the two cones:

V = Vcone (A) + Vcone (B)

= 1/3 × √(r2 – x2r2) × π x2r2 + 1/3 × √[r2 – (1-x)2 r2] × π (1-x)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 – (1-x)2 ] ×π (1-x)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.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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