Lesson 10.2: Stationary Points and Constrained Optimisation

Welcome to Lesson 10.2! 🎉 In this lesson, we will explore the fascinating world of stationary points and how to optimise functions subject to constraints using the method of Lagrange multipliers.

Learning Objectives

By the end of this lesson, you should be able to:

• Locate and classify stationary points of a function of two variables.
• Optimise subject to a constraint (the method of Lagrange multipliers).
• Model an optimisation problem from a worded context.
• Find and classify stationary points of a two-variable function.
• Solve a constrained optimisation problem using Lagrange multipliers.

Introduction

Optimisation is a key concept in mathematics, particularly when dealing with problems that require us to find the best possible solution under certain constraints. Whether it's maximising profit or minimising cost, knowing how to navigate these challenges is essential!

What are Stationary Points?

Stationary points occur when the first derivative of a function equals zero. In the case of functions of two variables, we extend this idea. For a function $f(x, y)$, a stationary point $(x_0, y_0)$ can be found where

$$\frac{\partial f}{\partial x}(x_0, y_0) = 0$$

and

$$\frac{\partial f}{\partial y}(x_0, y_0) = 0.$$

These points are crucial for determining local maxima, local minima, or saddle points of the function, which we will learn how to classify.

Classifying Stationary Points

To classify stationary points, we use the second derivative test. We calculate the second derivatives of the function:

• $f_{xx} = \frac{\partial^2 f}{\partial x^2}$
• $f_{yy} = \frac{\partial^2 f}{\partial y^2}$
• $f_{xy} = \frac{\partial^2 f}{\partial x \partial y}$

The Hessian Matrix

The Hessian matrix is defined as:

$$H = egin{bmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy} \end{bmatrix}$$

The determinant of the Hessian, denoted as $D$, is calculated as:

$$D = f_{xx} f_{yy} - (f_{xy})^2$$

Classification Criteria

1. If $D > 0$ and $f_{xx} > 0$, then $(x_0, y_0)$ is a local minimum.
2. If $D > 0$ and $f_{xx} < 0$, then $(x_0, y_0)$ is a local maximum.
3. If $D < 0$, then $(x_0, y_0)$ is a saddle point.
4. If $D = 0$, the test is inconclusive.

Example 1: Finding and Classifying Stationary Points

Consider the function:

$$f(x, y) = x^2 + y^2 - 4x - 6y$$

Step 1: Find the partial derivatives.

$$\frac{\partial f}{\partial x} = 2x - 4$$

$$\frac{\partial f}{\partial y} = 2y - 6$$

Step 2: Set the partial derivatives to zero.

$$2x - 4 = 0 \Rightarrow x = 2$$

$$2y - 6 = 0 \Rightarrow y = 3$$

Thus, we have the stationary point $(2, 3)$.

Step 3: Calculate second derivatives.

$$f_{xx} = 2, \quad f_{yy} = 2, \quad f_{xy} = 0$$

Step 4: Calculate $D$.

$$D = f_{xx} f_{yy} - (f_{xy})^2 = 2 \cdot 2 - 0^2 = 4$$

Since $D > 0$ and $f_{xx} > 0$, the point $(2, 3)$ is a local minimum. 🎯

Constrained Optimisation with Lagrange Multipliers

Sometimes, you need to optimise a function subject to a constraint. This is where Lagrange multipliers come into play!

The Method of Lagrange Multipliers

To optimise a function $f(x, y)$ subject to a constraint $g(x, y) = 0$, we introduce a new variable, $\lambda$ (the Lagrange multiplier), and solve the system:

$$

\frac{\partial f}{\partial x} = $\lambda$ \frac{\partial g}{\partial x} \

\frac{\partial f}{\partial y} = $\lambda$ \frac{\partial g}{\partial y} \

$g(x, y) = 0$

$$

Example 2: Using Lagrange Multipliers

Let’s maximise $f(x, y) = xy$ subject to the constraint $g(x, y) = x + y - 10 = 0$.

Step 1: Set up the equations using Lagrange multipliers.

$$\frac{\partial f}{\partial x} = y = \lambda \cdot 1 \quad (1)$$

$$\frac{\partial f}{\partial y} = x = \lambda \cdot 1 \quad (2)$$

$$g(x, y) = x + y - 10 = 0 \quad (3)$$

Step 2: Solve the system. From (1) and (2): $y = \lambda$, $x = \lambda$. Substituting into (3):

$$\lambda + \lambda - 10 = 0 \Rightarrow 2\lambda = 10 \Rightarrow \lambda = 5$$

Thus, $x = 5$ and $y = 5$.

Step 3: Find the maximum value.

$$f(5, 5) = 5 \cdot 5 = 25$$

Therefore, the maximum value of $f(x, y)$ subject to the constraint is 25! 🎉

Conclusion

In this lesson, we learned how to locate and classify stationary points of functions of two variables and how to apply the method of Lagrange multipliers to solve constrained optimisation problems. These concepts are fundamental tools for tackling real-world problems in economics, engineering, and data science.

Study Notes

• Stationary points occur where the first derivatives equal zero.
• Use the second derivative test to classify stationary points via the Hessian matrix.
• The method of Lagrange multipliers helps optimise a function subject to constraints.
• Always check the conditions of the Hessian to determine maxima, minima, or saddle points.
• Practice these techniques with different functions to become proficient in optimisation!