Maximum/Minimum Problems

Autore:Ku, Yin Bon (Albert)

Local and absolute extremum In

, we define an open disk centered at

with radius

, denoted by

, to be the set of all points such that their distance from

is smaller than

. Then we have the following definition: Definition: Suppose

be a function of two variables.

has a local maximum (minimum) at

for all

in the domain of

in an open disk

for some

. A local extremum of

means it is either a local maximum or a local minimum. The intuitive meaning of local maximum (minimum) of

is that

attains the greatest (smallest) value of

among all other points in the domain "close enough" to

. In the applet below, you can drag the point

to where you think a local maximum (minimum) occurs. Then verify it by the radius choosing

small enough such that the value of

at all the points in the yellow open disk with radius

are not greater (smaller) than that at

Definition:

has an absolute maximum (minimum) at

(

) for all

in the domain of

An absolute extremum of

means that it is either an absolute maximum or an absolute minimum. An absolute maximum (minimum) of

is necessarily a local maximum (minimum) of

. However, the converse is not true in general i.e. a local maximum (minimum) might not be an absolute maximum (minimum). Many real life problems can be formulated as optimization problems i.e. problems of finding absolute maximum/minimum of a certain function. However, not every function has an absolute maximum or minimum. For functions of one variable, we have extremum value theorem - a continuous function defined on a closed interval

attains its absolute maximum and minimum. For functions of two variables, we have the following analogous theorem: Theorem: Suppose

is a function of two variable such that it is continuous on a closed and bounded domain

(Note:

is bounded if it is contained in a an open disk with some radius.

is closed if it contains all its "boundary"). Then

attains its absolute maximum and minimum in

. Proof: Omitted.

Critical points In order to locate all local extrema of a function, we need the following theorem: Theorem: Let

be a function of two variables. If

has a local maximum or minimum at

and both

and

exist at

, then

. Proof: Define

i.e. fix

and regard

as a single-variable function of

. Then

has a local extremum at

i.e.

, which means that

. Similarly, define

i.e fix

and regard

as a single-variable function of

. The

has a local extremum at

i.e.

, which means that

. The above theorem is the reason why we have the following definition: Definition: Suppose

is in the domain such that

is contained in the domain of

for some

. Then

is a critical point of

if one of the following conditions holds:

At least one of and does not exist at

(Note: The reason why it is required that an open disk

is contained in the domain of

is because we want to make sure that

is in the "interior" of the domain i.e. not on the boundary of the domain.) Then the theorem says that if

is a local extremum of

, then it is a critical point of

. Therefore, when we are searching for local maximum or minimum in the interior of the domain, we only need to look among the critical points. The converse of the theorem is generally false - A critical point of

might not be a local extremum of

. Example: Let

. Then

and

. Therefore,

is a critical point of

because

. You can see from the applet below that

is actually a local minimum of

. Example: Let

. Then

and

. Therefore,

is a critical point of

because

. You can change the function to

in the applet below to see that

is actually a local maximum of

. Example: Let

. Then

and

. Therefore,

is a critical point of

because

and

do not exist. You can change the function to

in the applet below to see that

is actually a local minimum of

. Example: Let

. Then

and

. Therefore,

is a critical point of

because

. You can change the function to

in the applet below to see that

is neither a local maximum nor a local minimum of

. (Note: This kind of critical point is called a saddle point. More precisely, a critical point

is a saddle point if in every open disk

, there are points

for which

and points for which

The following is the main theorem for classifying critical points: Theorem (Second Derivative Test): Suppose

is a function of two variables such that all the second partial derivatives i.e.

are continuous on an open disk centered at

and

is a critical point of

i.e.

. Define

Then we have the following

If and , then has a local minimum at [\*]
If and , then has a local maximum at [\*]
If , has a saddle point at
If , then the test is inconclusive.

Example: Suppose

. Classify all critical points of

. Answer: First of all, compute all partial derivatives and set them to zero:

Combining the above equations and eliminate

, we get

Hence,

. For each value of

, we compute the corresponding value of

. Hence the critical points of

are

and

. To classify the critical points, we need to compute the second partial derivatives:

. For

Therefore,

is a saddle point of

. For

and

Therefore,

is a local maximum of

. For

and

Therefore,

is a local maximum of

. You can use the applet above to visualize all the critical points of

Exercise: Let . Classify all critical points of .

Finding absolute maximum/minimum Suppose

is continuous on a closed and bounded domain

. By extremum value theorem,

attains its absolute maximum and minimum in

. If absolute maximum/minimum value is attained in the interior of

i.e. not on the boundary, then it must be a local maximum/minimum, which implies that it is a critical point of

. Therefore, if we search for absolute maximum/minimum, we only need to check the values of

at the critical points and on the boundary of

. The general procedure for finding absolute maximum/minimum is as follows:

Find all critical points in and determine their values of .
Parametrize the boundary of and find the maximum and minimum values of on the boundary.
The greatest value found in the first two steps is the absolute maximum value of on , and the smallest value found in the first two steps is the absolute minimum value of on .

Example: Let

, which is a closed and bounded set. Find its absolute maximum and minimum values. Answer: Step 1: Find all critical points of

Moreover,

is in

. Hence

is the only critical point of

Step 2: Find maximum and minimum on the boundary. Since the boundary of

is a circle centered at

with radius 2, we can parametrize it by

for

. Then we have

. Compute the values of

at endpoints and the above two critical values:

Therefore, the maximum value on the boundary is

and the minimum value on the boundary is

. Step 3: Compare the values found in the previous steps.

(Absolute minimum value)

(Absolute maximum value)

Exercise: Suppose is defined on the domain , a closed region whose boundary is the triangle with vertices , and . Let be the line segment joining and , be the line segment joining and , and be the line segment joining and . (a) Parametrize , , and . (b) Find the absolute maximum and minimum values of on . (The applet below is the graph of with the domain .)

A real life application Suppose we want to made a rectangular box with open top using cardboard and its volume must be equal to

. Find the dimension of this box such that the least amount of cardboard is used. We need to formulate the above problem as an optimization problem - finding the absolute minimum of a function on a domain. First we let the dimension of the box, breadth x width x height, be

(all in ft). Let

be the total area of cardboard used for making the box. Then we have

. Combine them to eliminate

, we have

with

. And we need to minimize

among the domain

. Since the domain is not closed and bounded, we do not know if the absolute minimum is attained in the domain. First we find all critical point(s) of

Combine the above, we get

Therefore,

is the only critical point of

and

. In other words, the absolute minimum value, if exists, must be equal or less than 48. Let's analyze the function

. Suppose

and

cannot be too small because of the terms

and

. Also,

cannot be too large. More precisely,

implies the following inequalities:

The set

of all points satisfying these three inequalities is a closed and bounded set, as shown in the applet below. Since

on the the boundary of

, absolute minimum cannot be attained on the boundary. Hence the critical point

must be the point at which the absolute minimum of

is attained. The required dimensions of the box (breadth x width x height) is

. (

when

)

Maximum/Minimum Problems

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