• If you are citizen of an European Union member nation, you may not use this service unless you are at least 16 years old.

• Whenever you search in PBworks, Dokkio Sidebar (from the makers of PBworks) will run the same search in your Drive, Dropbox, OneDrive, Gmail, and Slack. Now you can find what you're looking for wherever it lives. Try Dokkio Sidebar for free.

View

# Optimization,-both-absolute-(global)-and-relative-(local)-extrema

last edited by 13 years, 11 months ago

Relative (aka Local) Extrema:

Before I go into the details of what an extrema is, I think a definition of an extrema would be a nice way to start.

Extrema- a maximum or minimum value of a function

1. The first step to calculating the relative extrema is calculating the derivative of a function .
2. After calculating the derivative (), set the derivative equal to 0.  Local extrema all have a slope of zero.  Setting the derivative, which is the slope at a specific point, equal to zero shows all points where the graph has a slope of zero, and as a result is an extrema.
3. Solve for x.
4. After solving for x, you must do one of two tests.
1. Option 1: The First Derivative Test

A) Draw a number line with the x value(s): <-----x1-------x2-----xn----->

B) Use numbers before and after the calculated x value, plugged into the first derivative equation,

to determine whether the numbers before and after are positive or negative.

C) If the number before is positive and the number after is negative,

the extrema is a local maximum.

D) If the number before is negative and the number after is positive,

the extrema is a local minimum.

E) On the occasion the number before and after is the same sign (positive or negative)

the number is not a local maximum or minimum.

F) The reason the first derivative test is true is because the first derivative is a slope the function.

The point where the slope changes from positive to negative is a maximum because the

function increases as much as it possibly can before its slope equals zero (critical point)

and then decreases after this point because the slope is then negative.

The opposite is true for a minimum.

2. Option 2: The Second Derivative Test

A) The first and most important step of the Second Derivative Test is to calculate

the second derivative.

B) Once the second derivative is calculated, plug in the values of x calculated from setting

the first derivative equal to zero.

C) If the value of the second derivative for the values of x is negative,

the point is a relative maximum because the graph is concave down.

A max must come at the top of a hill, and a hill like this is concave down.

D) If the value of the second derivative for the values of x is positive,

the point is a relative minimum because the graph is concave up.

A max must come at the bottom of a dip, and a dip like this is concave up.

E) On the occasion that the second derivative is equal to zero, the point is neither a relative

maximum or minimum, it is a point of inflection.  Inflection points represent a change in

concavity of the graph and do not represent a maximum or a minimum.

Absolute (aka Global) Extrema:

1. The first thing that must be done to calculate the absolute or global extrema is to first calculate the local extrema.
2. The maximums and or minimums calculated above only determine the maximums and minimums inside of the function but do not take into account end points.
3. After calculating the local maximums and minimums one must determine the starting and ending point of a function that has defined variables.
4. There is a possibility that either of these points can be higher than the local maximum or lower than the local minimum.  If this is the case then these points are the global extrema
5. If the end points are neither higher than the highest local maximum or lower than the lowest minimum, then the local extrema are also the global extrema.
6. Make sure to pick the highest or lowest local extrema if there are more than.

Differences between local and absolute extrema:

1. There is only one absolute extrema, one max and one minimum, where there can be multiple local extrema
2. The local extrema do not take into acount the beginning and ending points of a function.
3. The slope at local extrema is always zero (dy/dx = 0) where as the absolute extrema, because it can be a start or end point, can have a different slope.

Applet demonstrationg optimization: http://www.calculusapplets.com/boxproblem.

Just incase you want a graphical example of local and global exrema (maximums)

Now for some questions!!!!!!!!!! :-)

Multiple Choice Questions:

1. If f(x)= x3+x, then
1. $\inline \sqrt{3}/3$ =x is a local maximum of f
2. $\inline \sqrt{3}/3$ =x is a local minimum of f
3. $\inline \sqrt{3}$ =x is a local maximum of f
4. $\inline \sqrt{3}$ =x is a local minimum of f
5. -$\inline \sqrt{3}$ =x is a local minimum of f
2. How many relative or absolute maxima does x-cosx have on interval (-2$\inline \pi$, 2$\inline \pi$)
1. 1
2. 2
3. 3
4. 4
5. 5
3. Let f(x)= (x2-3)2.  The local minimum of f '(x) is
1. 7.99
2. 5.80
3. 3.25
4. -7.99
5. 6.28

Free Response Questions:

1. Function f(x)=2x+x
1. Find the local maximum of f(x).
2. Find the local minimum of f(x).
3. Evaluate (disclaimer: this does not relate to my topic but give it a try, refer to other wiki pages if neccessary

2: (0,0)

3: 2.25

2. A. Determine the constants a and b in order for the function f(x) = x3+ax2+c to have a relative minimum at x = 4 and a point of inflection at x = 1

B. Find the relative maximum of the function found in (A) after plugging in values of a and b, provided that f(0) = 1.

(A)

f(x) = 3x3+2ax2+b since f has a relative minimum at x=4 then

f '(4) = 3(4)2+2a(4) + b=0

=> 48 + 8a + b=0  (also check if f ''>0, and it is)

Because f has an inflection point at x = 1

=> f ''(1) =0 but f ''(x) =6x+2a

f ''(1)=0 => 6+2a = 0

=> a = -3

Substitute equaiton (2) into equaiton (1) to find b:

48 + 8(-3) + b=0

24+b = 0

b = -24

(B)

When we plug in a = -3 and b = -24 into

f(x) = x3+ax2+c we get

f(x) = x3+3x2+24x+c

But, f(0) = 1

=> 1 = f(0) = 0- 3(0)2-24(0)+c = c

Hence, f(x) = x3+3x2+24x+1

=> f'(x) = 3x2-6x-24

So,      f '(x) = 0

=> 3x2-6x-24 = 0

=> x2-2x-8 = 0 (multipying both sides by 1/3)

=> (x-4)(x+2) = 0

=> x = 4 or x = -2

Moreover, f ''(x) = 6x-6

Plugging in the citical points x = 4, x = -2, we see that

=> f ''(4) = 18 > 0 and f '' (-2) = -18 < 0

Hence, f has a relative maximum at x = -2 and a relative minimum at x = 4

So, the realtive maximum of is is

f(-2) = -8 - 12 + 48 + 1

= 29

3. y = f(x) is a function fhwere f ' and f '' exist and have the following characteristics:

 x x<-2 x=-2 -22 f ' (x) + 0 - - 0 + f '' (x) - - - + + +

If f(-2) = 8, f(0) = 4 and f(2) = 0, then

(A) Find all inflection points of -2ff

(B) Find all relative minimum and relative maximum values of -2f

(C) Discuss the concavity of -2f

(A) d2/dx2 (-2f) = -2f ''

From the chart, -2f '' > 0 in (-infinity, 0)

-2f '' < 0 in (0, infinity)

Therefore, -2f has inflection point at x=0

(B)

(i) d/dx (-2f) = -2f ' < 0 in (-infinity, -2) from the chart

-2f '' < 0 in (-2, infinity) from the chart

=> -2f has a relative minimum at x = -2

(ii) By similar argument as in (i), -2f has a relative maximum at x=2 and equals 0

(C)

Again from the table,  -2f '' > 0 in (-infinity, 0) and

-2f '' < 0 in (0, infinity)

So, it is concave upward in (-infinity,0) and concave downward in (0,infinity)

Bibliograhy:

Brook, Donald E., Donna M. Smith, and Tefera Worku. Mathematics Calculus AB. 1st. West Piscataway: Research and Education Association, 2000.

http://www.calculusapplets.com/boxproblem.html

http://dictionary.reference.com/search?q=extremum&db=luna

http://img.sparknotes.com/figures/E/eeb0fa64da5c2a1115cdf73c3ef2f0fc/derivapps1.gif

http://www.sosmath.com/calculus/diff/der14/der14.html