Maxima and Minima
The derivative measures the slope of the tangent
to a curve.
At the maximum or minimum points, the tangent
is horizontal and has slope zero.
Setting the derivative to zero, enables us to locate
these max/min points.
This process tells us where these max/min points
are but not whether they are a maximum or
minimum.
So how can we tell ??
1
y
y  0
y  0
x1
x
x2
The first derivative would give us the value of x1 and x2
but not the fact that x1 is a minimum point and x2 is a
maximum point.
We can establish this by taking the second derivative,
i.e. calculating the derivative of the derivative.
2
Test :
If y  0 , at a turning point (max/min point)
then the point is a minimum.
If y  0 , the point is maximum.
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Example: y  4 x 2  5 x  7
We know this has a maximum since a  4  0
y  8 x  5
Setting y  0 to find any turning points :
0  8 x  5  5  8 x
turning point
x5
8
y  8
Since -8 < 0 then this point is a maximum.
4
Example: y  x 2  5 ln x
 x  2x  5 x
y  2 x  5 1
Turning points occur when y  0
 0  2x  5
 5  2x
x
x
Since ln x is
5  2x2  x2  5
2
undefined when
x < 0, then 1.581 is
x   5  1.581
2
the only solution
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y  x 2  5 ln x
y  2 x  5  2 x  5 x 1
x
Now, y  2  5 x 2
When x = 1.581
y  2  51.581  4
2
Since 4 > 0, the point is a minimum.
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Note: If y  0, the point might not be a
maximum or minimum.
y  0
y  0
point of
inflection
y  0
y  0
y  0
y  0
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Solving Verbal Problems:
(1) Read the question carefully.
(2) Create an equation which gives a
mathematical description of the process.
(Sometimes given).
(3) Calculate the derivative and find turning
points.
(4) Establish whether or not turning points are
maximum or minimum or neither.
(5) Interpret and explain the results in
sentences.
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Note : (3)
If using the derivative does not
produce the desired result, then the
result will be one of the end points.
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e.g. looking for the minimum value of some function
The first derivative
will identify the
maximum value (B)
because it is a
turning point.
A
B
C
We are searching for the min value so check the
value of the function at A and C. The lesser of the
two gives the desired result.
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Example:
The cost per hour C (in dollars) of operating an
automobile is given by
C  0.12s  0.0012s 2  0.08, 0  s  60
where s is the speed in miles per hour. At what
speed is the cost per hour a minimum?
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Given C  0.12s  0.0012s 2  0.08
C   0.12  0.0024s
Setting C  0 gives
0  0.12  0.0024s
0.12
s
 50 s = 50 is a turning point
0.0024
C  0.0024  0
:. The turning point is a maximum.
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Check endpoints of s.
0  s  60
C 0  0.120  0.00120  0.08
 0.08
2
C 60  0.1260  0.001260  0.08
 2.96
The minimum value occurs at s = 0.
2
Costs are minimised when speed = 0 mi/hr.
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Example:
For a monopolist, the cost per unit of producing
a product is $3 and the demand equation is
p  10 q . What price will give the greatest
profit?
10
p
q
C  $3 per unit
14
Profit = total revenue - total costs
P  pq  3q
10
P
.q  3q
q
since
q1
q
1
q
1
2
2
P  10 q  3q
1
P  10q  3q
2
 P  5q
1
2
3
15
P  5q
1
2
3
Setting P  0
0  5q
1
2
3
1
3
q 2
5
2
 
 3   q 12 2
 
 5
25 

q  2.778   
 9
16
q = 2.778 is a turning point
1
3
P  5q  3  P  2.5q 2
3
When q  2.778, P  2.52.778 2
 0.54  0
:. q = 2.778 is a maximum point
The price that generates the maximum profit is
10
10
p

6
q
2.778
A price of $6 for the product will produce
maximum profit.
2
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Example
The cost of operating a truck on a throughway
(excluding the salary of the driver) is
0.11+(s/300) dollars per mile, where s is the
(steady) speed of the truck in miles per hour.
The truck driver’s salary is $12 per hour.
At what speed should the truck driver operate
the truck to make a 700-mile trip most
economical?
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s 

Cost per mile   0.11 

300 

:. For 700 miles,
s  ---------(1)

Cost  700 0.11 

300 

Driver’s salary = $12/hr, Since distance =
700
speed  time
number of hours =
s
700  8400 ----------(2)

:. Cost of driver  12

s
 s 
19
Total cost = (1) + (2)
s  8400

C  700 0.11 

300 
s

7 8400
C  77  s 
3
s
7
C    8400s 2
3
20
Setting C  0
7
0   8400s 2
3
7
2
8400s 
3
7
2
s 
3  8400
1
2
1
7


2
s  2  

 3  8400 
s  60
Turning point
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Checking with C 
C  16800s 3
s  60, C  1680060
 0.078  0
3
:. The turning point is a minimum.
So costs are minimised when s = 60 mi/hr
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A final note :
It is worthwhile checking end points of the variables
domain even when the calculus has produced a result.
A
B
Calculus will reveal a maximum point at A but it can
be seen the max value of the function occurs at B
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