Lesson 8.4: Stationary Points and Simple Optimisation

Introduction

In this lesson, we will explore the concept of stationary points and how they relate to optimisation problems using polynomial functions. Understanding stationary points is crucial because they help us identify where a function reaches its maximum or minimum values, which is a key concept in both pure mathematics and practical applications such as physics and economics.

Learning Objectives

• Finding stationary points by setting the derivative to zero.
• Classifying maxima and minima for simple polynomial functions.
• Solving straightforward optimisation problems from worded contexts.
• Locating the stationary points of a polynomial function.
• Classifying a stationary point as a maximum or minimum.

Understanding Stationary Points

What are Stationary Points?

A stationary point of a function occurs where the derivative of the function is equal to zero or is undefined. The derivative measures the rate of change (slope) of the function at a given point. When this rate of change is zero, it suggests that the function is neither increasing nor decreasing, indicating a potential maximum, minimum, or a point of inflection.

Finding Stationary Points

To find stationary points for a polynomial function, we follow these steps:

1. Differentiate the function: Find the first derivative of the function, $ f'(x) $.
2. Set the derivative to zero: Solve the equation $ f'(x) = 0 $ for $ x $.
3. Find corresponding $ y $-values: Substitute the $ x $-values back into the original function to find $ y $-coordinates.

Example: Finding Stationary Points

Let's consider the polynomial function:

$$\displaystyle f(x) = x^3 - 3x^2 + 4$$

1. Differentiate the function: We first need to find $ f'(x) $.

$ f'(x) = 3x^2 - 6x $

1. Set the derivative to zero: Solve $ 3x^2 - 6x = 0 $.

$ 3x(x - 2) = 0 $

This gives us two solutions: $ x = 0 $ and $ x = 2 $.

1. Find corresponding $ y $-values: Now we substitute these $ x $-values back into the original function:

• For $ x = 0 $:

$ f(0) = 0^3 - 3(0)^2 + 4 = 4 $

• For $ x = 2 $:

$ f(2) = 2^3 - 3(2)^2 + 4 = 8 - 12 + 4 = 0 $

Thus, the stationary points are $ (0, 4) $ and $ (2, 0) $.

Classifying Stationary Points

Second Derivative Test

To classify whether a stationary point is a maximum, minimum, or neither, we use the second derivative test. This involves:

1. Finding the second derivative, $ f''(x) $.
2. Evaluating $ f''(x) $ at each stationary point:

• If $ f''(x) > 0 $: the point is a local minimum.
• If $ f''(x) < 0 $: the point is a local maximum.
• If $ f''(x) = 0 $: the test is inconclusive.

Example: Classifying the Stationary Points

Continuing from our previous example with:

$$\displaystyle f'(x) = 3x^2 - 6x$$

1. Find the second derivative:

$ f''(x) = 6x - 6 $

1. Evaluate at stationary points:

• For $ x = 0 $:

$ f''(0) = 6(0) - 6 = -6 $

Since $ f''(0) < 0 $, $ (0, 4) $ is a local maximum.

• For $ x = 2 $:

$ f''(2) = 6(2) - 6 = 6 $

Since $ f''(2) > 0 $, $ (2, 0) $ is a local minimum.

Optimisation Problems

Solving Optimisation Problems

Optimisation often entails finding the maximum or minimum value of a real-world scenario that can be modeled by a function. This involves:

1. Formulating the problem into a mathematical function.
2. Finding the stationary points of the function.
3. Classifying the stationary points to identify the optimum solution.

Example: Optimisation Problem

Problem: A farmer wants to fence a rectangular area for a garden. The area must be 100 square meters. What dimensions should be used to minimize the amount of fencing required?

1. Let the length be $ l $ and the width be $ w $**. The area constraint gives us:

$ A = l \times w = 100 \ \text{m}^2 $(1)

1. The perimeter $ P $ (amount of fencing needed) is given by:

$ P = 2l + 2w $ (2)

1. Express $ w $ in terms of $ l $** using equation (1):

$$ w = \frac{100}{l} $$

1. Substituting $ w $ into $ P $** from equation (2):

$$ P = 2l + 2\left(\frac{100}{l}

ight) = 2l + $\frac{200}{l}$ $$

1. Now find the derivative of $ P $ with respect to $ l $:

$$ P'(l) = 2 - \frac{200}{l^2} $$

1. Set the derivative equal to zero and solve:

$$ 2 - \frac{200}{l^2} = 0 $$

Rearranging gives:

$$ \frac{200}{l^2} = 2 $$

$$ l^2 = 100 $$

$$ l = 10 \ \text{m} $$

1. Finding $ w $ using equation (1):

$$ w = \frac{100}{10} = 10 \ \text{m} $$

1. Thus, the dimensions that minimize the fencing required are $ 10 \ \text{m} \times 10 \ \text{m} $, yielding a square garden.

Conclusion

In this lesson, we introduced the concept of stationary points and how to find and classify them using polynomial functions. Understanding how to find and classify maximums and minimums is essential for solving optimisation problems that arise in various practical contexts. By mastering these skills, students will build a solid foundation for tackling more advanced topics in calculus.

Study Notes

• A stationary point occurs where the derivative equals zero or is undefined.
• To find stationary points of a function, differentiate, set the derivative to zero, and solve for $ x $.
• Classify stationary points using the second derivative test:
• $ f''(x) > 0 $: local minimum.
• $ f''(x) < 0 $: local maximum.
• $ f''(x) = 0 $: inconclusive.
• For optimisation problems, formulate the function, find stationary points, and classify them to determine optimum solutions.