Applications of Maximisation and Minimisation Revision Notes for HSC SSCE Mathematics Advanced

Applications of Maximisation and Minimisation

Introduction to real-world optimisation

In everyday life and industry, we often need to find the best possible outcome for a given situation. Calculus provides powerful tools to solve these optimisation problems. Some common applications include:

Maximising the volume of a box constructed from a sheet of cardboard
Minimising fuel consumption during a flight
Maximising profits from manufacturing and selling products
Minimising the amount of material used in product packaging

These problems can be solved using differentiation techniques, but only when we can establish a clear mathematical relationship between the variables involved. The key is transforming a real-world scenario into a mathematical function that we can analyse.

chatImportant

Critical Principle: Whenever you claim that a stationary point represents a maximum or minimum value, you must always justify this claim through proper mathematical analysis. Simply finding where the derivative equals zero is not enough.

The four-step problem-solving process

When tackling maximisation and minimisation problems, follow this systematic approach:

Step 1: Define your variables

Clearly identify and label the quantities involved. Typically, you will need to:

Let $y$ (or another appropriate letter) represent the quantity you want to maximise or minimise
Let $x$ (or another appropriate letter) represent the quantity that can be varied (the independent variable)

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Drawing a diagram is usually helpful at this stage to visualise the problem and understand the relationships between variables.

Step 2: Form the equation

Create an equation connecting your two variables, making sure to note any restrictions on the domain. These restrictions might come from:

Physical constraints (e.g., lengths cannot be negative)
Practical limitations (e.g., a box cannot have zero dimensions)
Given conditions in the problem

Step 3: Find the global maximum or minimum

Use calculus techniques to locate and verify the global maximum or minimum:

Differentiate the function with respect to the independent variable
Find stationary points by setting the first derivative equal to zero
Use the second derivative test or first derivative test to determine the nature of each stationary point
Check boundary values if the domain is restricted
Identify which stationary point gives the global maximum or minimum

Step 4: Write a clear conclusion

State your answer clearly, including:

The optimal value of the quantity being maximised or minimised
The values of all relevant variables at this optimal point
Units where applicable

Worked example: Maximising box volume

This example demonstrates how to maximise the volume of an open box created by cutting squares from the corners of a flat sheet.

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Worked Example: Maximising the Volume of an Open Box

Problem: An open rectangular box is to be made by cutting square corners from a square piece of cardboard measuring 60 cm × 60 cm, then folding up the sides. What is the maximum volume of the box, and what are its dimensions?

Solution:

Step 1: Define variables and understand the setup.

Let $V$ be the volume of the box (this is what we want to maximise).

Let $x$ be the side length of each square that is cut from the corners.

When we fold up the sides, the box will have:

Height: $x$ cm
Base: a square with side length $(60 - 2x)$ cm (because we remove $x$ from each end)

Step 2: Form the volume equation.

The volume of a rectangular box is length × width × height:

$V = x(60 - 2x)^2$

Expanding this expression:

V = x(3600 - 240x + 4x^2) V = 3600x - 240x^2 + 4x^3

The domain is $0 \leq x \leq 30$ (we cannot cut squares larger than half the cardboard's side length).

Step 3: Find the maximum value.

Differentiate with respect to $x$ :

V' = 3600 - 480x + 12x^2

Factor this expression:

V' = 12(300 - 40x + x^2) V' = 12(x - 30)(x - 10)

Set $V' = 0$ to find stationary points:

$12(x - 30)(x - 10) = 0$

Therefore $x = 10$ or $x = 30$

These are the only stationary points, with no discontinuities in the domain.

Now use the second derivative test to determine the nature of each stationary point:

$V'' = -480 + 24x$

At $x = 10$ :

$V''(10) = -480 + 240 = -240 < 0$

This indicates a local maximum (the curve is concave down).

At $x = 30$ :

$V''(30) = -480 + 720 = 240 > 0$

This indicates a local minimum (the curve is concave up).

Calculate the volume at $x = 10$ :

$V(10) = 10(60 - 20)^2 = 10(40)^2 = 10 \times 1600 = 16000 \text{ cm}^3$

Since this is the only local maximum in the domain $0 \leq x \leq 30$ , the point $(10, 16000)$ gives the global maximum.

Step 4: State the conclusion.

The maximum volume is 16 000 cm³, achieved when the box has dimensions 10 cm × 40 cm × 40 cm (height × length × width).

Worked example: Minimising surface area

This example shows how to minimise the amount of material needed to construct a cylindrical container with a fixed volume.

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Worked Example: Minimising the Surface Area of a Cylindrical Can

Problem: A cylindrical soft drink can must have a volume of 250 cm³.

a) Show that the height of the can is $h = \frac{250}{\pi r^2}$ , where $r$ is the base radius.

b) Show that the total surface area is $S = 2\pi r^2 + \frac{500}{r}$ .

c) Show that $r = \frac{5}{\pi^{\frac{1}{3}}}$ gives a global minimum of $S$ in the domain $r > 0$ .

d) Show that to minimise the surface area, the diameter of the base should equal the height.

Solution:

Part a: Establish the relationship between height and radius.

Let the height of the can be $h$ cm.

The volume of a cylinder is:

$\text{Volume} = \pi r^2 h$

Since the volume must be 250 cm³:

$250 = \pi r^2 h$

Solving for $h$ :

$h = \frac{250}{\pi r^2}$

Part b: Derive the surface area formula.

The total surface area consists of:

Two circular ends, each with area $\pi r^2$
One curved side with area $2\pi rh$

Therefore:

$S = 2\pi r^2 + 2\pi rh$

Substitute the expression for $h$ from part a:

S = 2\pi r^2 + 2\pi r \times \frac{250}{\pi r^2} S = 2\pi r^2 + \frac{500}{r}

where $r > 0$

Part c: Find and verify the minimum surface area.

Differentiate $S$ with respect to $r$ :

\frac{dS}{dr} = 4\pi r - \frac{500}{r^2}

Rewrite with a common denominator:

\frac{dS}{dr} = \frac{4\pi r^3 - 500}{r^2}

Set $\frac{dS}{dr} = 0$ to find stationary points:

4\pi r^3 - 500 = 0 4\pi r^3 = 500 r^3 = \frac{125}{\pi} r = \frac{5}{\pi^{\frac{1}{3}}}

Now apply the second derivative test:

$\frac{d^2S}{dr^2} = 4\pi + \frac{1000}{r^3}$

Since $r > 0$ , both terms are positive, so $\frac{d^2S}{dr^2} > 0$ for all $r > 0$ .

This confirms that the stationary point is a local minimum. Since it's the only stationary point in the domain $r > 0$ , it must be the global minimum.

Part d: Verify the diameter equals height relationship.

When $r = \frac{5}{\pi^{\frac{1}{3}}}$ :

h = \frac{250}{\pi r^2} h = \frac{250}{\pi} \times \frac{\pi^{\frac{2}{3}}}{25} h = \frac{250 \times \pi^{\frac{2}{3}}}{25\pi} h = \frac{10}{\pi^{\frac{1}{3}}} h = 2r

Therefore, the diameter $(2r)$ equals the height when the surface area is minimised. This is an elegant result showing that the most efficient cylindrical can has equal height and diameter.

Cost and time optimisation problems

Many real-world problems involve finding the optimal speed or rate to minimise costs. There are often competing factors to balance:

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Understanding the Trade-offs:

At slow speeds:

Wages and fixed costs increase (because the journey takes longer)
Time-dependent expenses accumulate

At high speeds:

Fuel consumption increases (often proportional to speed squared)
Wear and tear on equipment increases
Operating costs rise

The optimal speed lies somewhere between these extremes, where the total cost is minimised. If we can express the cost as a mathematical function of speed, calculus enables us to find this optimal value.

Worked example: Minimising travel costs

This example illustrates how to find the most economical speed for a journey.

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Worked Example: Finding the Optimal Speed to Minimise Travel Costs

Problem: The cost $C$ (in dollars per hour) of running a boat depends on its speed $v$ km/h according to the formula:

$C = 500 + 40v + 5v^2$

a) Show that the total cost for a 100 km trip is $T = \frac{50000}{v} + 4000 + 500v$ .

b) What speed will minimise the total cost of the trip?

Solution:

Part a: Derive the total cost formula.

The fundamental relationship is:

$\text{Time} = \frac{\text{distance}}{\text{speed}}$

For a 100 km journey at speed $v$ km/h:

$\text{Time} = \frac{100}{v} \text{ hours}$

The total cost equals the cost per hour multiplied by the time taken:

T = (\text{cost per hour}) \times (\text{time for trip}) T = (500 + 40v + 5v^2) \times \frac{100}{v} T = \frac{50000}{v} + 4000 + 500v

where $v > 0$

Part b: Find the speed that minimises total cost.

Differentiate $T$ with respect to $v$ :

\frac{dT}{dv} = -\frac{50000}{v^2} + 500

Rewrite with a common denominator:

\frac{dT}{dv} = \frac{-50000 + 500v^2}{v^2} \frac{dT}{dv} = \frac{500(-100 + v^2)}{v^2}

Factor the numerator:

$\frac{dT}{dv} = \frac{500(v - 10)(v + 10)}{v^2}$

Set $\frac{dT}{dv} = 0$ :

$500(v - 10)(v + 10) = 0$

This gives $v = 10$ or $v = -10$

Since $v > 0$ (speed must be positive), the only relevant stationary point is $v = 10$ . There are no discontinuities in the domain.

Apply the second derivative test:

$\frac{d^2T}{dv^2} = \frac{100000}{v^3}$

For all $v > 0$ , we have $\frac{d^2T}{dv^2} > 0$ , confirming that $v = 10$ gives a local minimum.

Since this is the only stationary point in the domain $v > 0$ , it must be the global minimum.

Conclusion: A speed of 10 km/h will minimise the cost of the trip.

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This result makes intuitive sense: at this speed, the savings from reduced time (lower fixed costs) exactly balance the additional expenses from increased fuel consumption and operating costs.

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Key Points to Remember:

Always justify your answer: When claiming a stationary point is a maximum or minimum, you must use the second derivative test or another valid method to verify this.
Follow the four-step process: Define variables, form equations with restrictions, find the global optimum, and write a clear conclusion.
Check the domain: Physical constraints often limit the possible values. Always identify and state these restrictions.
Use the second derivative test: If $f''(x) < 0$ at a stationary point, it's a local maximum. If $f''(x) > 0$ , it's a local minimum.
Include units in your final answer: Make sure your conclusion states the optimal value with appropriate units (cm³, km/h, dollars, etc.).

Applications of Maximisation and Minimisation (HSC SSCE Mathematics Advanced): Revision Notes