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Lecture 11
Max-Min Problems
Maxima and Minima
Problems of type:
“find the largest, smallest, fastest, greatest, least, loudest, oldest,
youngest, cheapest, most expensive, … etc.
If problem can be phrased as in terms of finding the largest value of a
function then one is looking for the highest and lowest points on a graph
(if they exist)
There is not a one-step test to detect
“highest” and lowest points on a graph.
What we can do is detect where relative
maxima and minima occur.
Relative extrema occur at
critical numbers (remember
that end points are critical
numbers)
Strategy for solving Max-Min:
• Phrase problem as a function for which one is to
find largest or smallest value
• Find all critical numbers of the function (including
end points)
• Make a table of values of the function at the critical
numbers
• IF the function is defined on a closed interval then
the largest (smallest) functional value in the table
is the maximum (minimum) value of the function.
• If not on a closed interval can still detect the local
maxima and minima – may be able to determine if
one of them is an absolute max or min by
sketching the graph.
length = x
length = f(x) = 8 x
3
x
2
0
Max Value
P(x) = 2 x + 2( 8 x 3 )
P(x) 2 x 16 2 x
P ' (x) 2 6 x
0 x 2
3
2
Critical numbers = 0,2,
1
3
3
Function is on closed interval [0,2]









1

3
x



16



4



4

3  16
9

P(x)
0
2
3
Want x-value of highest point
2
f ' 3 x  2 x
Critical numbers = -1 (ep), 0 (ep), -2/3
Function is defined on closed interval so
max value at critical number is max value of
function.
=
x ( 3 x 2 )
 x


 0


 -1


 -2

3
f(x) 


0 


0 


4 

27 
MAX
Length = f(x) – g(x) or g(x)-f(x)
h(x) = f(x) – g(x)
3
2
h( x ) x  2 x  3 2 x
x
-1.5 x 1.5
2
h ' (x) 3 x  4 x 2
2
0 3 x  4 x 2
2 1
x 
10
3 3
x 1.720759220
or
2 1
x 
10
3 3
(> 1.5) or
x -.3874258863
Critical numbers = -1.5 (ep), 1.5 (ep),
Just because a
derivative is 0 does not
mean it is a critical number
of the function under
consideration – Here
1.720359.. is not in the
domain.
2
3

10
3
 x


 -1.5


 1.5


  .38742
h (x) 


-1.875 


-1.125 


3.4165
MAX
Length y
Length x
2
vol x y
2
10 x y
y
10
x
Area = base + 4 sides = x*x + 4*x*y
2
Area = base + 4 sides = x*x + 4*x*y
2 4 x 10
A( x ) x 
2
x
2 40
A( x ) x 
x
A ' (x) 2 x
40
x
A ' (x) 2
Looking for lowest point –
will occur at x = critical number
near 3.
2
A( 2.71441 ) 22.1042
3
x  20
x
A ‘ = 0 if x =
2
20
 1
 
 3
=
2.7144
This is an example where
the domain is not a closed
Interval but we can still
determine that the min
occurs at the one critical
number
You can buy any amount of motor oil at $.50 per quart.
At $1.10 you can sell 1000 quarts but for each penny
increase in the selling price you will sell 25 fewer quarts.
Your fixed costs are $100 regardless of how many quarts
you sell. At what price should you sell oil in order to maximize
your profit. What will be your maximum possible profit?
Profit = Income - Costs
= (number sold)*(selling price) –
[ (number purchased)*(purchase price)
+ fixed costs]
Let x = increase in price in pennies
Profit = (1000 -25*x)(1.10+.01*x) - [ (1000-25*x)*.50 + 100]
p( x ) 500.00 985.00 x 25 x
2
p ' (x) 500.00 5.00 x .25 x
2
p ' (x)  5.00 .50 x
0 x  40
Critical numbers = 0 (ep), 40 (ep), -5/.5 = 10
 x


 0


 40


  10
p(x) 


500.00


-100.0 


525.00
Maximum profit
We determine here
that there is no need
to consider x > 40 which
gives us a closed interval
to work on
Optimum selling price = $1.10 + (-.01*10) = $1