Maxima and Minima in Practical Context

Using differentiation to solve real-world optimisation problems

SPEC 7.3 • APPLICATIONS OF DIFFERENTIATION

What is Optimisation?

Definition

Optimisation is the process of finding the maximum or minimum value of a quantity, subject to given constraints.

In practical problems, we often need to find:

Maximum area, volume, profit, efficiency

Minimum cost, material, distance, time

Key Insight

At a maximum or minimum, the derivative equals zero. This is the mathematical foundation of all optimisation problems.

Real-World Applications

Engineers design containers to minimise material. Farmers maximise field area with limited fencing. Businesses optimise production to maximise profit.

The 6-Step Method

Follow this systematic approach for every optimisation problem:

1
IDENTIFY the quantity to be optimised (maximised or minimised)
2
DEFINE variables and write an expression for the quantity
3
FIND constraints - relationships between variables from the problem
4
SUBSTITUTE to express the quantity in terms of ONE variable only
5
DIFFERENTIATE and solve $f^{'} (x) = 0$ to find stationary points
6
VERIFY using second derivative and check answer makes sense in context

Important

Step 4 is crucial - you cannot differentiate with two variables. Always eliminate one variable first!

Worked Example - Open-Top Box01

A square sheet of card (24 cm × 24 cm) has squares of side x cm cut from each corner. It is folded to make an open-top box. Find x for maximum volume.

1

After cutting, base = (24 - 2x) × (24 - 2x), height = x

2

Volume: V = x(24 - 2x)² = x(576 - 96x + 4x²)

3

Expand: V = 4x³ - 96x² + 576x

4

Differentiate:

\frac{d V}{d x} = 12 x^{2} - 192 x + 576

5

Factor: 12(x² - 16x + 48) = 12(x - 4)(x - 12) = 0

6

Solutions: x = 4 or x = 12

7

Domain check: 0 < x < 12, so x = 4 is valid (x = 12 gives zero base)

8

Second derivative:

\frac{d ^{2} V}{d x ^{2}} = 24 x - 192

, at x = 4: 96 - 192 = -96 < 0 → Maximum

Answer: Maximum volume = 4 × 16² = 1024 cm³ when x = 4 cm

Worked Example - Cylinder02

A closed cylinder has total surface area 600π cm². Find the radius that gives maximum volume.

1

Surface area: 2πr² + 2πrh = 600π

2

Divide by 2π: r² + rh = 300, so rh = 300 - r², hence h = (300 - r²)/r

3

Volume: V = πr²h = πr² × (300 - r²)/r

4

Simplify: V = πr(300 - r²) = 300πr - πr³

5

Differentiate:

\frac{d V}{d r} = 300 π - 3 π r^{2}

6

Set to zero: 300π - 3πr² = 0 → r² = 100 → r = 10

7

Second derivative:

\frac{d ^{2} V}{d r ^{2}} = - 6 π r

, at r = 10: -60π < 0 → Maximum

8

Height: h = (300 - 100)/10 = 20 cm

Answer: Maximum volume when r = 10 cm and h = 20 cm

Worked Example - Cone03

A cone has fixed slant height l = 10 cm. Find the radius that maximises the volume of the cone.

1

Constraint: l² = r² + h², so 100 = r² + h², giving h = √(100 - r²)

2

Volume:

V = \frac{1}{3} π r^{2} h = \frac{1}{3} π r^{2} 100 - r^{2}

3

Since V > 0, maximising V is equivalent to maximising V²

4

V^{2} = \frac{π ^{2}}{9} r^{4} (100 - r^{2}) = \frac{π ^{2}}{9} (100 r^{4} - r^{6})

5

Differentiate:

\frac{d ( V ^{2} )}{d r} = \frac{π ^{2}}{9} (400 r^{3} - 6 r^{5}) = \frac{2 π ^{2} r ^{3}}{9} (200 - 3 r^{2})

6

Set to zero: Since r ≠ 0, we need 200 - 3r² = 0 → r² = 200/3

7

Therefore: r = √(200/3) = 10√(2/3) ≈ 8.16 cm

8

Height: h = √(100 - 200/3) = √(100/3) = 10/√3 ≈ 5.77 cm

Answer: Maximum volume when r ≈ 8.16 cm and h ≈ 5.77 cm

Worked Example - Closest Point04

Find the point on the curve y = x² that is closest to the point (3, 0).

1

Distance from (x, y) to (3, 0): D = √[(x - 3)² + y²]

2

Since y = x², substitute: D = √[(x - 3)² + x⁴]

3

Minimise D² instead (easier): f(x) = (x - 3)² + x⁴

4

Expand: f(x) = x² - 6x + 9 + x⁴ = x⁴ + x² - 6x + 9

5

Differentiate: f'(x) = 4x³ + 2x - 6

6

Factor: 2(2x³ + x - 3) = 2(x - 1)(2x² + 2x + 3) = 0

7

x = 1 (quadratic 2x² + 2x + 3 has no real roots as discriminant = 4 - 24 < 0)

8

When x = 1, y = 1² = 1

Answer: The closest point is (1, 1)

Worked Example - Cost Function05

A company's production cost is $C = x^{2} + \frac{2000}{x}$ pounds for x items. Find the number of items that minimises total cost.

1

Cost function: C = x² + 2000/x = x² + 2000x⁻¹

2

Differentiate:

\frac{d C}{d x} = 2 x - \frac{2000}{x ^{2}}

3

Set to zero: 2x - 2000/x² = 0

4

Multiply by x²: 2x³ - 2000 = 0 → x³ = 1000

5

Solve: x = 10

6

Second derivative:

\frac{d ^{2} C}{d x ^{2}} = 2 + \frac{4000}{x ^{3}}

7

At x = 10: 2 + 4000/1000 = 2 + 4 = 6 > 0 → Minimum

8

Minimum cost: C = 100 + 200 = £300

Answer: Minimum cost of £300 when producing 10 items

Practice

Question 1: A farmer has 200 m of fencing to create a rectangular pen against an existing barn (so only 3 sides need fencing). Find the maximum possible area of the pen.

1

Let width = x m, length (along barn) = y m

2

Constraint: 2x + y = 200 → y = 200 - 2x

3

Area: A = xy = x(200 - 2x) = 200x - 2x²

4

dA/dx = 200 - 4x = 0 → x = 50

5

d²A/dx² = -4 < 0 → Maximum

6

When x = 50, y = 100

Answer: Maximum area = 50 × 100 = 5000 m²

Practice

Question 2: A rectangular box with a square base and no top has total surface area 300 cm². Find the dimensions that maximise the volume.

1

Let base side = x cm, height = h cm

2

Surface area: x² + 4xh = 300 → h = (300 - x²)/(4x)

3

Volume: V = x²h = x² × (300 - x²)/(4x) = x(300 - x²)/4

4

Simplify: V = (300x - x³)/4 = 75x - x³/4

5

dV/dx = 75 - 3x²/4 = 0 → x² = 100 → x = 10

6

d²V/dx² = -3x/2, at x = 10: -15 < 0 → Maximum

7

Height: h = (300 - 100)/40 = 5 cm

Answer: Maximum volume = 10 × 10 × 5 = 500 cm³ when base = 10 cm, height = 5 cm

Practice

Question 3: The cost of running a machine is £

(\frac{50}{v} + 2 v)

per hour, where v is the speed in m/s. Find the speed that minimises the running cost.

1

Cost function: C = 50/v + 2v = 50v⁻¹ + 2v

2

Differentiate: dC/dv = -50/v² + 2

3

Set to zero: -50/v² + 2 = 0

4

Rearrange: 2 = 50/v² → v² = 25 → v = 5

5

Second derivative: d²C/dv² = 100/v³

6

At v = 5: 100/125 = 0.8 > 0 → Minimum

7

Minimum cost: C = 50/5 + 2(5) = 10 + 10 = £20 per hour

Answer: Minimum cost of £20/hour when speed = 5 m/s

Summary

The 6-Step Method

StepAction

1. IdentifyWhat quantity to maximise/minimise?

2. DefineWrite expression for the quantity

3. ConstraintFind the relationship between variables

4. SubstituteEliminate one variable

5. DifferentiateSolve f'(x) = 0

6. VerifyUse f''(x) to confirm max/min

Key Exam Skills

Draw a diagram and label all variables

Write the constraint equation before substituting

Check your solution is within the valid domain

Always verify with the second derivative test

Give your final answer in context with units