Maximum and Minimum Values

Maximum and Minimum Values

Consider a function

We say that is a

Problem: Find extrema (maximum or minimum) of on some domain D.

Local maximum if for in D “around”

Absolute maximum if for all in D

Local minimum if for in D “around”

Absolute minimum if for all in D

Maximum and Minimum ValuesTheorem: If has a local maximum or minimum at a point , and the first-order partial derivatives of exist there, then and

Geometrically:

If the function f has a local maximum (or minimum) at , then the tangent plane must be horizontal with equation .

Since the equation of the tangent plane is

it follows that

is a necessary but not sufficient condition for a maximum or a minimum. We could have a saddle point at (a,b)

Maximum and Minimum Values

NOTES:

( , ) ( , ) 0 x yf a b f a b or or both may not exist at a local maximum or minimum.

( , )xf a b ( , )yf a b A maximum or minimum could be on the boundary of the domain D.

Maximum and Minimum Values

Let be a point in the interior of the domain D.

We say is a critical point for if either

• and , or

• or or both do NOT exist at

Example 1: Find the critical points of

The partial derivatives are equal to 0, when and so the only critical point is

Critical points are candidates for maximum and minimum.

From the figure we can see that is a minimum. The surface is the elliptic paraboloid withvertex at

Maximum and Minimum Values

How do we determine analytically whether a critical point is a maximum, a minimum or neither?

Second Derivative Test:

We define the Discriminant: 2( , ) ( , ) ( , ) [ ( , )] xx yy xyD x y f x y f x y f x y

Assume the second partial derivatives of are continuous “around” and

Classification of

+ + Local Minimum

+ − Local Maximum

− Saddle point

0 Test is inconclusive

Maximum and Minimum Values – Example 2 Find and classify the critical points of

( , ) 6 , ( , ) 6 , ( , ) 3 xx yy xyx y x f x y y f x yf

Critical point

Discriminant Type

Saddle since

Local minimum since and

Setting the partial derivatives equal to 0, yields and

Substitute from the first equation into the second equation. This gives with roots and

The two critical points are and (1,1)

Contour plot

Maximum and Minimum Values - Application

Find the point (x, y, z) on the plane z = 4x + 3y + 3 which is closest to the origin.

We will minimize the square of the distance from the point (x, y, z) to the origin, ubject to the constraint z = 4x + 3y + 3

Substituting z into yields

Find the critical points:

Solving the system gives 6 9,

13 26 x y

Substituting into the equation of the plane gives 326

z

The point on the plane closest to the origin is 6 9 3, ,

13 26 26

Maximum and Minimum Values

ABSOLUTE MAXIMUM AND MINIMUM VALUES

EXTREME VALUE THEOREM: If is continuous on a closed, bounded region D in R2, then f attains an absolute maximum and an absolute minimum at some points and in D.

To find absolute maximum and minimum:

Step 1. Find the values of f at the critical points

Step 2. Find the extreme value of f on the boundary of DStep 3. The largest value in 1. and 2. is the absolute maximum, the smallest value is the absolute minimum.

The boundary consists of the line segment and the parabola. We first check the line segment.

, (5,25)

Step1: Find the critical points:

Maximum and Minimum Values - Example 3Find the absolute maximum and minimum of the function

on the region D bounded by y = x2 and y = 36.

Evaluate the function: f(5, 25) = 0

Step 2: Find the extreme values of on the boundary of D

This is an increasing function of x, so its minimum value is and its maximum value is

Substituting y = 36 into the expression for gives:

Now let’s check the boundary . Substituting into the expression for , gives

Maximum and Minimum Values - Example 3 continued

We can see from the graph of that the function attains its minimum

( 6,36) 121 f5 25, 148.153 9

f

Step 3: Compare all the values from Step 1 and 2.

Absolute minimum

Absolute maximum

To find the maximum we solve The roots are and

Values of :