Maximising the Volume of a Cone inside a Sphere

Assignment Brief

Engineering Mathematics 1

1. DEMONSTRATE AN UNDERSTANDING OF THE BASIC MATHEMATICAL TECHNIQUES REQUIRED BY ENGINEERS.

2. APPLY THE MATHEMATICAL TECHNIQUES REQUIRED TO SOLVE PROBLEMS.

3. DEVELOP AN UNDERSTANDING OF CALCULUS AND APPLY CALCULUS TO SIMPLE ENGINEERING PROBLEMS

The MOD are proposing to house a radar system inside a geodesic dome. The dome is in the form of a sphere of diameter 42m, part of which will be below ground. One option being considered is that the radar system will be housed inside a right circular cone inside the sphere as shown in the diagram below: Before making a decision, it is necessary to determine the dimensions of the cone that maximises the available volume.

1. If h is the height of the cone, show that the volume of the cone in terms of h is given by `

2. Using calculus determine the value of h that maximises the volume of the cone [3 marks]

3. Using calculus methods show that the value you have obtained in (ii) is a maximum [2 marks]

4. Determine the radius of the base and the volume of the cone

The formula for the period, T, of a pendulum of length depends on its length L. For small oscillation the formula is 𝑇𝑇 = 2 𝜋𝜋 �𝐿𝐿 𝑔𝑔 . For larger oscillations the formula for the period T is given by the following integral: 𝑇𝑇 = 4 �𝐿𝐿 𝑔𝑔 ∫ 𝑑𝑑𝑑𝑑 �1−𝑘𝑘2 sin2(𝑥𝑥) 𝜋𝜋/2 0 where 𝑘𝑘 depends on the angle from which the pendulum is released. Complete the following table using k = 0.8: x 0.0 𝜋𝜋 12 𝜋𝜋 6 𝜋𝜋 4 𝜋𝜋 3 5 𝜋𝜋 12 𝜋𝜋 2 1 �1 − 𝑘𝑘2sin2(𝑥𝑥) For a pendulum, of length 1.5m, estimate the period of the pendulum (to 6dp) using (i) Trapezium rule and (ii) Simpson’s rule [10 marks]

The surface of an air intake is machined according to the function 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 6 𝑥𝑥3 + 3 𝑦𝑦2 − 8 𝑥𝑥 + 2 𝑦𝑦 + 1 part of which is shown below. Find and classify the locations of the stationary points of this function.

Sample Answer

Maximising the Volume of a Cone inside a Sphere

The sphere has a diameter of 42 m, so its radius R = 21 m.

A cone is inscribed inside the sphere with its apex at the bottom and base touching the inner surface of the sphere.
Let the height of the cone be h and the base radius be r.

Expressing Volume in terms of h

The general equation of the sphere (taking the centre as the origin) is:

x² + y² + z² = R²

If the apex of the cone is at z = –R and the base at z = –R + h,
then the radius of the cone at the base is given by:

r² = R² – (R – h)²
r² = R² – (R² – 2Rh + h²)
r² = 2Rh – h²

The volume of a cone is:

V = (1/3) π r² h

Substitute r²:

V = (1/3) π h (2Rh – h²)

Since R = 21,

V = (1/3) π (42h² – h³)

Finding h that maximises the volume

Differentiate V with respect to h:

dV/dh = (1/3) π (84h – 3h²)
Simplify:

dV/dh = π h (28 – h)

For stationary points, set dV/dh = 0:

π h (28 – h) = 0

So h = 0 or h = 28.
(We ignore h = 0 since that gives zero volume.)

Thus, h = 28 m gives the stationary point.

Show it is a Maximum

Take the second derivative:

d²V/dh² = π (28 – 2h)

Substitute h = 28:

d²V/dh² = π (28 – 56) = –28π

Since the second derivative is negative, this is a maximum.

Radius of Base and Maximum Volume

We already know r² = 2Rh – h².

Substitute R = 21 and h = 28:

r² = (2 × 21 × 28) – 28²
r² = 1176 – 784
r² = 392

r = √392 = 19.8 m (approximately)

Now, calculate the maximum volume:

Vmax = (1/3) π r² h
Vmax = (1/3) π (392)(28)
Vmax = (1/3) π (10976)
Vmax = 3658.67 π
Vmax ≈ 11,494.0 m³

Hence, the cone of maximum volume has:
Height h = 28 m
Radius r = 19.8 m
Maximum Volume V = 11,494 m³

Continued...