Part 6 - Maximum and minimum values

Introduction

In this part we’ll cover how we find maximum and minimum values for a function, we’ll also see how this is related to the 2nd derivative.

Maximum and minimum values

Let’s first recall how we define maximum and minimum values for functions of one variable.

Definition 1 (Local and absolute extrema)

For maximum:

A function of one variable has local maximum at $a$ , if $f(a) \geq f(x)$ when $x$ is near $a$ .

For minimum: A function of one variable has local minimum at $a$ , if $f(a) \leq f(x)$ when $x$ is near $a$ .

Similarly, for two variables: A function of two variables has local maximum at $(a, b)$ if $f(a, b) \geq f(x, y)$ when $(x, y)$ is near $(a, b)$ .

A function of two variables has local minimum at $(a, b)$ if $f(a, b) \leq f(x, y)$ when $(x, y)$ is near $(a, b)$ .

For absolute max/min, we define them as: If $f(a, b) \geq f(x, y)$ for any $(x, y) \in D$ , we have an absolute maximum.

If $f(a, b) \leq f(x, y)$ for any $(x, y) \in D$ , we have an absolute minimum.

How do we find these minimums and maximums?

How to find min/max

Let’s recall for functions of one variable, min/max points have derivate of zero, therefore:

f'(a) = 0 \ | \ f'(a) \ \exists

For two variables, this is the same:

Theorem 1

If a function of two variables has local min/max at $(a, b)$ , then $f_x(a, b) = 0$ and $f_y(a, b) = 0$ given that $f_x(a, b)$ and $f_y(a, b)$ exists.

Proof

Let $(a, b)$ be a point of local maximum and $f_x(a, b)$ and $f_y(a, b)$ exists. $f(a, b) \geq f(x, y)$ when $(x, y)$ near $(a, b)$ .

In particular, take y = b:

f(a, b) = f(x, b) \ | \ x \text{ near } a

Let $g(x) = f(x, b)$ , this means:

g(a) = f(a, b)

We can say that $a$ is a point of local maximum:

g'(a) = 0

Which is just the same as:

f_x(a, b) = 0

Similarly, for $f_y(a, b) = 0$ , we just choose to fix $x = a$ instead.

Geometrical sense

In functions of one variable, at points of local min/max, the tangent line is horizontal to our point, if the line exists.

For functions of two variables, instead of a line, we have a plane. The tangent plane is horizontal to our point, if the plane exists.

Let’s properly define what we’ve covered so far:

Definition 2 (Critical point)

$(a, b)$ is a critical point if $\nabla f(a, b) = 0 \textbf{ or } \nabla f(a, b) \ \nexists$

In other words, if $(a, b)$ is a critical point of local minimum/maximum then it is a critical point.

However, the other way around this implication is not true. A critical point might not be a local minimum/maxium.

Example 1

Find the critical point(s) for the function:

f(x, y) = x^2 + y^2 - 2x - 6y + 14

Let’s use our definition:

\nabla f = 0 \newline \langle 2x - 2, 2y - 6 \rangle = 0

We find that $x = 1$ and $y = 3$ . This means $(1, 3)$ is a critical point (and the only one!).

But how do we actually know if a critical point is a local minimum or maximum?

2nd derivate test

For functions of one variable, if we do the 2nd derivate test:

Suppose $a$ is a point and $f''(a)$ exists, then:

f''(a) > 0 \ | \ \text{local min} \newline f''(a) < 0 \ | \ \text{local max} \newline f''(a) = 0 \ | \ \text{does not say anything}

Let’s see what we can do in functions of two variables.

Theorem 2 (2nd derivate test for functions of two variables)

Suppose $(a, b)$ is a critical point of a function, $f$ , of two variables. Suppose 2nd order partial derivatives exists and are continous.

Let $D = D(a, b) = f_{xx}(a, b) f_{yy}(a,b) - \left(f_{xy} (a,b)\right)^2$

Then:

D > 0 \text{ and } f_{xx}(a, b) > 0 \ | \ \text{local minimum} \newline D > 0 \text{ and } f_{xx}(a, b) < 0 \ | \ \text{local maximum} \newline D < 0 \ | \ \text{neither, this is a saddle point} \newline

Example 2

Let $f(x, y) = x^2 - y^2$ Find critical point(s):

\nabla f = 0 \newline \lange 2x, -2y \rangle = 0

This means $(0, 0)$ is a critical point:

f_{xx} = 2 f_{yy} = -2 f_{xy} = 0

This means:

D = -4 \ | \ \text{saddle point}

Let’s take one last example

Example 3

Find local min/max/saddle points, $f(x, y) = x^4 + y^4 - 4xy + 1$

\nabla f = 0 \langle 4x^3 - 4y, 4y³ - 4x \rangle = 0

\begin{cases} 4x^3 - 4y = 0 \newline 4y^3 - 4x = 0 \end{cases}

Solving this equation system we find that:

\begin{cases} x = 0 \newline x = \pm 1 \end{cases}

\begin{cases} y = 0 \newline y = \pm 1 \end{cases}

Let’s now find the 2nd order partial derivates:

f_{xx} = 12x^2 \newline f_{yy} = 12y^2 \newline f_{xy} = f_{yx} = -4

D = 144x^2y^2 - 16

\begin{cases} D(0, 0) = -16 \ | \ \text{saddle point} \newline D(1, 1) = 128 \ | \ \text{local minimum} \newline D(-1, -1) = 128 \ | \ \text{local minimum} \newline \end{cases}