maxima and minima
DESCRIPTION
Differential calculusTRANSCRIPT
Maxima and Minima Problems
Absolute Extrema
In such a case, the number is called the absolute maximum
value of f on I.
Definition. A function f has an absolute maximum value on an interval I if there is some number c in I such that
for each x in I. xfcf
f c
In such a case, the number is called the absolute minimum
value of f on I.
Definition. A function f has an absolute minimum value on an interval I if there is some number c in I such that
for each x in I. f c f x
f c
Definition. An absolute extremum of a function f on an interval I is either an absolute maximum value or an absolute minimum value of f on I.
Illustration: Determine the absolute extrema of the functions represented by the following graphs.
The absolute minimum of the function occurs at b.
The absolute maximum of the function occurs at a.
The absolute minimum of the function occurs at a.
The absolute maximum of the function occurs at b.
Figure 4.4.1(d)
The absolute maximum of the function occurs at c.
The absolute minimum of the function occurs at a.
The absolute minimum of the function occurs at c.
The absolute maximum of the function occurs at b.
Theorem The Extreme-Value Theorem
If a function f is continuous on a closed interval [a,b] then f has an absolute maximum value and an absolute minimum value on [a,b].
A
B
C
D
E
F
G
Relative minimum pts:
C, E
Absolute minimum pt:
G
Relative maximum pts:
B, D,
Absolute maximum pt:
DF
1. Find the critical number(s) of f.
2. Find the value of f at each of the critical numbers of f on .
3. Find the values of and .
4. The largest of the values from steps 2 and 3 is the absolute maximum value of f; the smallest of the values is the absolute minimum value of f.
How do you find the absolute extrema of a continuous function f on the closed interval [a, b]?
b,a af bf
if .Find the absolute extrema of on
Example1
0513 xx
05143' 2 xxxf
1 or 53x x
f 2,0 xxxxf 57 23
Solution:
• Solve for the critical numbers of f on (0,2).
Does f have absolute maximum and minimum values?
Yes
Candidates for absolute extrema:
3 27 5 , 0,2f x x x x x
262 f
1 23
3 27f
2x
0x
3
1x
xf
00 f
Conclusion
f has an absolute minimum value of -23/27 at x = 1/3.
f has an absolute maximum value of 26 at x=2
3 27 5 , 0,2f x x x x x
(2,26)
(0.33,-0.85)
if .Find the absolute extrema of g on [-3,-1]
Example 2
2' 3 5 0g x x
3 5 4g x x x
Solution:
• Solve for the critical numbers of g on (-3,-1).
Does f have absolute maximum and minimum values? Yes
For what value/s of x will g’(x) be zero?
Thus, g has no critical number on [-3,-1].
None
Compute the function values of g at the endpoints.
3 5 4, 3, 1g x x x x
31 1 5 1 4
10
g
1x
3x 33 3 5 3 4
46
g
g has an absolute minimum value of -46 at x = -3.
g has an absolute maximum value of -10 at x = -1
Find the absolute extrema of the following functions on the indicated interval.
Exercise
2 on ( 2,3]f x x
4 on [2,5)f x xa.
b.
c.
d. on [ 1,2]2
xf x
x
3 on [ 3, )f x x
Steps in Solving Max-Min Problems
Step 1. Understand the problem.
Read the problem carefully. Identify the information you need to solve the problem. What is unknown? What is given? What is required?
Step 2. Develop a mathematical model of the problem.
Draw pictures and label the parts that are important to the problem. Introduce a variable to represent the quantity to be maximized or minimized. Write a function that relates the variable/s to the problem.
Step 3. Find the domain of the function.Determine what values of the variable
make sense in the problem.
Step 4. Identify the critical points and endpoints.
Step 5. Solve the mathematical model.
Step 6. Interpret the solution.
A piece of wire 10 feet long is cut into two pieces. One piece is bent into the shape of a circle and the other into the shape of the square. How should the wire be cut so that the combined area of the two figures is as small as possible?
Example 1
x x10
Total Area =Area of Circle + Area of Square
Total Length of Wire
= Circumference of Circle + Perimeter of Square
Solution:
4
10 xs
xs 104
2
xr
2
2
x
xr 2
2
4
10
x
r s
Total Area =
2
2
x2
4
10
x
Total Area =
16
20100
4
22 xxxxfA
where 0,10x
5' ' 0
2 4 8
x xA f x
10
4x
.4
10
44
x
10x
0x
2510 7.96f
2
10 125 100
4 4
3.5
f
250 6.25
4f
2
( )2
xf x
2
4
10
x
Total Area: , 0,10x
Absolute minimum
Absolute minimum value of f at x = 4.4 is 3.5
Thus, 4.4 ft and 5.6 ft of wire should be used to form the circle and the square, respectively, to have the smallest combined area of f(4.4) = 3.5 sq. ft.
x = 4.4
10 - x = 5.6
Theorem
Suppose a function f is continuous on an interval I containing the number c. If f(c) is a relative extremum of f on I and c is the only number in I for which f has a relative extremum, then f(c) is an absolute extremum of f on I.
Example 2
A farmer has 2400 feet of fencing materials and wants to fence off a rectangular field that borders a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area?
w
l
Let l be the length (in feet) of the
field parallel to the river
w be the width (in feet) of the
rectangular field
Solution:
If A is the area of the
rectangular field, then
wwwfA 22400
600w
lwA
24002 wl (length of fencing materials )
042400' wwf
22400 2 , 0,1200f w w w w
4'' wf and '' 600 4 0f Thus, f has a relative maximum at w = 600.
w
l
2400 2l w (length in terms of width )
Since the only relative maximum value of f is at w = 600, then the absolute maximum value of f occurs when w = 600.
Therefore, using 2400 feet of fencing materials the rectangular field must be 1200 ft by 600 ft to have the largest possible area of f(600) = 720,000 sq.ft.
w = 600
l = 2400 – 2w = 2400 – 2 (600)
= 1200
Example 3Find two numbers whose difference is
100 and whose product is a minimum.
100P f x x x 100x-yxyP
Solution:
Let x and y be the two numbers
P be the product of the two numbers
where 100y x
2 100 , ,f x x x x
2'' xf and '' 50 2 0f ' 2 100 0f x x
Hence, f has a relative minimum value at x = 50.
Since the only relative minimum of f is when x = 50, then f(50) is the absolute minimum value.
Therefore, the two numbers in the problem are 50 and - 50. The minimum product is f(50) = - 2500.
2 100 , ,f x x x x
50x
x = 50 y = x - 100 = - 50
Example 4Find the height of the right circular cylinder of maximum
volume which can be inscribed in a sphere of radius 9 cm.
2
814
hh
22 29
2
hr
2V r h
Solution:
Let h be the height (in cm) of the cylinder
22 81
4
hr
V(h) be the volume (in sq.cm) of the cylinder
h/29
r
r be the radius (in cm) of the cylinder
3181 , 0,18
4V h h h h
0 0V
3 408 .91 12V
23' 81 0
4V h h
V has an absolute maximum at h = 18.4
Therefore, the height of the right circular cone which can be inscribed in the sphere of radius 9 cm should be 18.4 cm to have a maximum volume of 3124.9 cm2.
108 18.4h
3181 , 0,18
4V h h h h
18 0V