Maximum and Minimum Problems Revision Notes for VCE SSCE Mathematical Methods

Maximum and Minimum Problems

Introduction to optimisation problems

In many real-world situations, we need to find the best possible outcome by making a quantity either as large or as small as possible. This process is called optimisation.

Maximisation means making a quantity as large as possible. Examples include:

Maximising profit from sales
Maximising attendance at an event
Maximising the area of an enclosure

Minimisation means making a quantity as small as possible. Examples include:

Minimising manufacturing costs
Minimising fuel consumption
Minimising time taken

We use differentiation to solve optimisation problems by finding where the derivative equals zero, which identifies potential maximum or minimum values.

General technique for solving maximum and minimum problems

Follow these systematic steps:

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Step 1: Define variables

Clearly identify what you're trying to maximise or minimise, and define appropriate variables.

Step 2: Write the equation

Express the quantity to be optimised as a function of a single variable. If there are multiple variables, use constraint equations to eliminate all but one.

Step 3: Differentiate

Find the derivative of your function.

Step 4: Find stationary points

Solve the equation where the derivative equals zero. These are your stationary points.

Step 5: Test the nature

Use a gradient chart (sign chart) to determine whether each stationary point is a maximum, minimum, or neither.

Step 6: Check practical constraints

Verify that your answer makes sense in the practical context. If the domain is restricted, also check the boundary values.

Worked example: Maximising area with fixed perimeter

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Worked Example: Maximising Area with Fixed Perimeter

Question: A farmer has sufficient fencing to make a rectangular pen of perimeter $200$ metres. What dimensions will give an enclosure of maximum area?

Solution:

Let the length of the rectangle be $x$ metres.

Then the width is $(100 - x)$ metres.

The area is $A$ m², where:

A = x(100 - x)

A = 100x - x^2

The maximum value of $A$ occurs when $\frac{dA}{dx} = 0$ .

\frac{dA}{dx} = 100 - 2x

$\frac{dA}{dx} = 0$ implies $x = 50$

From the gradient chart, the maximum area occurs when $x = 50$ .

The pen with maximum area has dimensions 50 m by 50 m, giving an area of 2500 m².

chatImportant

Key observation: The maximum area occurs when the rectangle is actually a square. This is a useful result to remember.

Worked example: Minimising with a constraint equation

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Worked Example: Minimising with a Constraint Equation

Question: Two variables $x$ and $y$ are such that $x^4y = 8$ . A third variable $z$ is defined by $z = x + y$ . Find the values of $x$ and $y$ that give $z$ a stationary value and show that this value of $z$ is a minimum.

Solution:

From the constraint equation $x^4y = 8$ , we can express $y$ in terms of $x$ :

y = 8x^{-4}

Substitute into the equation $z = x + y$ :

z = x + 8x^{-4}

Now $z$ is expressed as a function of one variable, $x$ . Differentiate:

\frac{dz}{dx} = 1 - 32x^{-5}

A stationary point occurs where $\frac{dz}{dx} = 0$ :

1 - 32x^{-5} = 0

32x^{-5} = 1

x^5 = 32

x = 2

When $x = 2$ , the corresponding value is $y = 8 \times 2^{-4} = \frac{1}{2}$ .

Substitute $x = 2$ into the equation for $z$ :

z = 2 + \frac{8}{16} = 2\frac{1}{2}

Table

The gradient chart confirms this is a minimum.

The minimum value of $z$ is 2½ and occurs when $x = 2$ and $y = \frac{1}{2}$ .

Worked example: Maximising volume with a surface area constraint

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Worked Example: Maximising Volume with Surface Area Constraint

Question: A cylindrical tin canister closed at both ends has a surface area of $100$ cm². Find, correct to two decimal places, the greatest volume it can have. If the radius of the canister can be at most $2$ cm, find the greatest volume it can have.

Solution:

Let the radius be $r$ cm, the height be $h$ cm, and the volume be $V$ cm³.

The surface area equation:

100 = 2\pi r^2 + 2\pi rh

(two circular ends plus curved surface)

The volume equation:

V = \pi r^2h

From the surface area equation, express $h$ in terms of $r$ :

h = \frac{1}{2\pi r}(100 - 2\pi r^2)

Substitute into the volume equation:

V = \pi r^2 \times \frac{1}{2\pi r}(100 - 2\pi r^2)

V = 50r - \pi r^3

A stationary point occurs when $\frac{dV}{dr} = 0$ :

\frac{dV}{dr} = 50 - 3\pi r^2

$\frac{dV}{dr} = 0$ implies $50 - 3\pi r^2 = 0$

r = \pm\sqrt{\frac{50}{3\pi}} \approx \pm 2.3

Since radius must be positive, $r \approx 2.3$ cm.

When $r = 2.3$ , $V \approx 76.78$ cm³.

The gradient chart confirms this is a maximum.

Part 2: If the radius can be at most $2$ cm, the domain is restricted to [0, 2].

We've shown that $\frac{dV}{dr} > 0$ for all $r \in [0, 2]$ . This means the volume is increasing throughout this interval.

Therefore, the maximum occurs at the boundary value $r = 2$ .

The maximum volume is $V = 100 - 8\pi \approx 74.87$ cm³.

chatImportant

Important principle: When working with restricted domains, always check both stationary points AND boundary values to find the absolute maximum or minimum.

Worked example: Maximising cross-sectional area using trigonometry

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Worked Example: Maximising Cross-Sectional Area Using Trigonometry

Question: The cross-section of a drain is to be an isosceles trapezium, with three sides of length $2$ metres, as shown. Find the angle $\theta$ that maximises the cross-sectional area, and find this maximum area.

Solution:

Let $A$ m² be the area of the trapezium.

The area formula is:

A = \frac{1}{2} \times \text{height} \times \text{(sum of parallel sides)}

From the diagram:

Height = $2\sin\theta$
Parallel sides are $2$ m and $(2 + 2 \times 2\cos\theta)$ m

A = \frac{1}{2} \times 2\sin\theta \times (2 + 2 + 4\cos\theta)

A = \sin\theta \cdot (4 + 4\cos\theta)

Differentiate using the product rule:

A'(\theta) = \cos\theta \cdot (4 + 4\cos\theta) - 4\sin^2\theta

A'(\theta) = 4\cos\theta + 4\cos^2\theta - 4(1 - \cos^2\theta)

A'(\theta) = 4\cos\theta + 8\cos^2\theta - 4

Set the derivative equal to zero:

8\cos^2\theta + 4\cos\theta - 4 = 0

2\cos^2\theta + \cos\theta - 1 = 0

(2\cos\theta - 1)(\cos\theta + 1) = 0

\cos\theta = \frac{1}{2} \text{ or } \cos\theta = -1

The practical restriction is $0 < \theta \leq \frac{\pi}{2}$ .

Therefore $\theta = \frac{\pi}{3}$ (which is $60°$ ).

When $\theta = \frac{\pi}{3}$ :

A = \sin\frac{\pi}{3}(4 + 4\cos\frac{\pi}{3})

A = \frac{\sqrt{3}}{2}(4 + 2) = 3\sqrt{3} \text{ m}^2

The maximum cross-sectional area is 3√3 m².

Maximum rates of increase and decrease

The derivative of a function tells us the instantaneous rate of change. By differentiating the derivative (finding the second derivative), we can determine when the rate of change itself is at a maximum or minimum.

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Remember:

If $\frac{dy}{dx} > 0$ , then $y$ is increasing as $x$ increases
If $\frac{dy}{dx} < 0$ , then $y$ is decreasing as $x$ increases

Worked example: Maximum rate of population change

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Worked Example: Maximum Rate of Population Change

Question: The number of bacteria in a culture at time $t$ is given by $N(t) = 36te^{-0.1t}$ .

a) Sketch the graphs of $N(t)$ against $t$ and $N'(t)$ against $t$ .

b) Find the maximum rates of increase and decrease of the population and the times at which these occur.

Solution:

Part a)

N(t) = 36te^{-0.1t}

N'(t) = 36e^{-0.1t} - 3.6te^{-0.1t} = e^{-0.1t}(36 - 3.6t)

Part b)

Let $R(t) = N'(t) = e^{-0.1t}(36 - 3.6t)$ represent the rate of change of the population.

From the graph, the maximum value of $R(t)$ occurs at $t = 0$ .

The maximum rate of increase is $R(0) = 36$ bacteria per unit time.

To find the maximum rate of decrease, we need to find the minimum value of $R(t)$ .

Differentiate $R(t)$ :

R'(t) = -7.2e^{-0.1t} + 0.36te^{-0.1t}

R'(t) = e^{-0.1t}(-7.2 + 0.36t)

$R'(t) = 0$ implies $t = 20$

At $t = 20$ :

R(20) = -36e^{-2} \approx -4.9

The maximum rate of decrease is approximately $4.9$ bacteria per unit time, occurring at $t = 20$ .

chatImportant

Exam tip: When asked about "maximum rate of decrease", you're looking for the most negative value of the derivative, which is the minimum of the rate function.

The second derivative and points of inflection

The second derivative of a function $f$ , denoted $f''(x)$ or $\frac{d^2y}{dx^2}$ , is found by differentiating the derivative.

The second derivative provides information about the concavity of a curve and helps identify points of inflection.

Concave up and concave down

Concave up: When $f''(x) > 0$ for all $x$ in an interval, the gradient is increasing. The curve is concave up (shaped like a cup or smile).

Concave down: When $f''(x) < 0$ for all $x$ in an interval, the gradient is decreasing. The curve is concave down (shaped like a cap or frown).

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Visual Memory Aid:

Concave up = Happy face (smile) = $f''(x) > 0$
Concave down = Sad face (frown) = $f''(x) < 0$

Points of inflection

A point of inflection is where a curve changes from concave up to concave down, or vice versa.

At a point of inflection of a twice differentiable function:

$f''(x) = 0$ (necessary condition)
There must be a change in concavity (also required)

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Important: Having $f''(x) = 0$ alone does not guarantee a point of inflection. You must verify that the concavity actually changes.

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Example: Finding a Point of Inflection

For $f(x) = 3x^3 - 4x + 1$ :

f'(x) = 9x^2 - 4

f''(x) = 18x

At $x = 0$ : $f''(0) = 0$

For $x < 0$ : $f''(x) < 0$ (concave down)

For $x > 0$ : $f''(x) > 0$ (concave up)

Since the concavity changes at $x = 0$ , there is a point of inflection at (0, 1).

Using gradient charts effectively

A gradient chart (or sign chart) is a powerful tool for determining the nature of stationary points. Here's how to construct one:

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Constructing a Gradient Chart:

Step 1: Draw a table with rows for $x$ , $\frac{df}{dx}$ , and shape of $f$ .

Step 2: Mark the critical point where $\frac{df}{dx} = 0$ .

Step 3: Test the sign of the derivative on either side of the critical point.

Step 4: Use these symbols:

$/$ for increasing
$\backslash$ for decreasing
$-$ for horizontal (at the stationary point)

Step 5: Interpret:

If the derivative changes from $+$ to $0$ to $-$ , you have a maximum
If the derivative changes from $-$ to $0$ to $+$ , you have a minimum

Remember!

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Key Points to Remember:

Optimisation problems involve finding maximum or minimum values using differentiation
Set the derivative equal to zero to find stationary points
Use gradient charts to determine whether stationary points are maxima or minima
For restricted domains, always check boundary values as well as stationary points
The second derivative indicates concavity: positive means concave up, negative means concave down
Points of inflection occur where concavity changes and $f''(x) = 0$
When solving practical problems, always verify your answer makes sense in the real-world context

Maximum and Minimum Problems (VCE SSCE Mathematical Methods): Revision Notes