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CHAPTER 3 ADDITIONAL APPLICATIONS OF THE DEREVATIVE
§3.1 Increasing and Decreasing Functions; Relative Extreme
P200 16, 21, 27, 29, 30, 32, 39, 41, 47, 55, 57, 61, 72
In problems 16 and 21 find the intervals of increase and decrease for the given function.
16. g (t ) 
1
1
 2
t  1 (t  1) 2
2
Solution (e.g. 3.1.2)
The function is defined for all real numbers t , and its derivative is
g (t )  
2t
4t
 2t (t 2  1)  4t  2t (t 2  1)



(t 2  1) 2 (t 2  1) 3
(t 2  1) 3
(t 2  1) 3
Which has g (t )  0 at t  1 , t  0 , and t  1 . Thus, there are four intervals on which the sign of
g (t ) does not change: namely, t<-1, -1<t<0, 0<t<1, and t>1. Choosing test numbers in these
intervals (say, -2, -0.5, 0.5 and 2, respectively), we find that
g (2)  0
g (0.5)  0
g (0.5)  0
g ( 2)  0
We conclude that g (t ) is increasing (graph rising) for t<-1 and for 0<t<1 and that it is decreasing
(graph falling) for -1<t<0 and for t>1.
-------
++++++
++
----
-1
21. f ( x)  x 
-------
++++++
0
++
1
----
t
1
x
Solution (e.g. 3.1.2, 3.1.5)
The function is defined for x  0 , and its derivative is
f ( x) 
1
2 x

1
2x x

x 1
2x x
So f (x) exits everywhere in the domain of the function f (x) , and f ( x)  0 only when x  1. Next,
mark the point on a number line restricted to the domain of f (x) (namely, x  0 ), and determine
the sign of the derivative f (x) on the subintervals 0  x  1 and x  1 to obtain the arrow
diagram. Choosing test numbers in these intervals (say, 0.5 and 2, respectively), we find that
f (0.5)  0 and f (2)  0
We conclude that f (x) is decreasing (graph rising) for 0  x  1 and that it is increasing (graph
falling) for x  1.
------0
++++++
1
----
x
++
In problems 29, 30 and 32 determine the critical numbers of the given function and classify each
critical point as a relative maximum, or neither.
t
29. f (t )  2
t 3
Solution (e.g. 3.1.3)
The function f (t ) is defined for all t , and its derivative is
f (t ) 
3t2
(t 2  3) 2
Since the derivative exists for all x , the only critical numbers are where f (t )  0 ; that is,
t  3 and t   3 which divide the x axis into three intervals, on each of which the sign of the
derivative does not change; namely, t   3 ,  3  t  3 and t  3 . Choose a test number c in
each of these intervals (say, -2, 0 and 2 respectively) and evaluate f (c ) in each case:
f (2)  0
f (0)  0
f (2)  0
Thus, the graph of g (x ) falls for t   3 and t  3 , and rises for  3  t  3 , so there must be a
relative maximum at t  3 and relative minimum at t   3 .
--------
++++++
 3
++
-------x
3
30. f (t )  t 9  t
Solution (e.g. 3.1.3, 3.1.5)
The function f (t ) is defined only for t  9 , and its derivative is
f (t )  9  t 
t
2 9t

3(6  t )
2 9t
So f (t ) does not exit at the endpoint t  9 of the function f (t ) , and f (t )  0 only when t  6 .
Thus, there are two intervals, on each of which the sign of the derivative does not change;
namely, t  6 and 6  t  9 . Choose a test number c in each of these intervals (say, 0 and 8
respectively) and evaluate f (c ) in each case:
f (0)  0 and f (8)  0
Thus, the graph of f (x) rises for t  6 , and falls for 6  t  9 , so there must be a relative maximum
at t  6 .
++++++
++
-------6
9
x
2 3

x x2
Solution (e.g. 3.1.3)
32. g ( x)  4 
The function g (x ) is defined only for t  0 , and its derivative is
g ( x) 
2
6 2( x  3)
 3 
2
x
x
x3
Since the derivative exists in the domain of function g (x ) , the only critical numbers are
where g ( x )  0 ; that is, x  3 . So x axis are divided into three intervals, on each of which the sign
of the derivative does not change; namely, x  0 , 0  x  3 and x  3 . Choose a test number c in
each of these intervals (say, -1, 1 and 4 respectively) and evaluate g (c) in each case:
g ( 1)  0 g (1)  0
g ( 4)  0
Thus, the graph of g (x ) rises for x  0 and x  3 , and falls for 0  x  3 , so there must be relative
minimum at x  3 .
++++++
++
-------0
++++++
3
++
x
In problems 39 and 41, use calculus to sketch the graph of the given function.
39. f (t )  2t 3  6t 2  6t  5
Solution (e.g. 3.1.4)
Since f (t ) is a polynomial, it is defined for all x. Its derivative is
f (t )  6t 2  12t  6  6(t  1) 2
Since the derivative exits for all x, the only critical number is where f (t )  0 ; namely, at t  1 ,
which divide the x axis into two intervals, on each of which the sign of the derivative does not
change; namely, t  1 and t  1 . Choose a test number c in each of these intervals (say, -2 and 1
respectively) and evaluate f (c ) in each case:
f (2)  0 and f (1)  0
Thus, the graph of f (t ) rises for both t  1 and t  1 , and has horizontal tangent where t is -1,
as indicated in this arrow diagram:
++++++
++
++++++
-1
++
x
Interpreting the diagram, the graph rises to a horizontal tangent at t  1 , then continuous rising
indefinitely. And we know f (1)  3 ,
t
t 3
Solution (e.g. 3.1.3, 3.1.4)
41. g (t ) 
2
The function f (t ) is defined for all t , and its derivative is
f (t ) 
3t2
(t 2  3) 2
Since the derivative exists for all x , the only critical numbers are where f (t )  0 ; that is,
t  3 and t   3 which divide the x axis into three intervals, on each of which the sign of the
derivative does not change; namely, t   3 ,  3  t  3 and t  3 . Choose a test number c in
each of these intervals (say, -2, 0 and 2 respectively) and evaluate f (c ) in each case:
f (2)  0 f (0)  0
f (2)  0
Thus, the graph of g (x ) falls for t   3 and t  3 , and rises for  3  t  3 , so there must be a
relative maximum at t  3 and relative minimum at t   3 .
--------
++++++
 3
++
-------x
3
Interpreting the diagram, the graph falls to the relative minimum at t   3 , the rises to the
relative maximum at t  3 , after which it falls indefinitely. We find that g ( 3 )  
3
, then we
6
can get the diagram
61. MORTGACE REFINANCING When interest rates are low, many homeowners take the
opportunity to refinance (vt. 再供给资金; 出售证券以清偿 (债务) 或付清 (债券等)) their mortgages
(n. 抵押, 抵押单据; 抵押所借的款项). As rates start to rise, there is often a flurry ((股票市场行情等)短
时间波动) of activity as latecomers (迟到者) rush in to refinance while they still can do so profitably.
Eventually, however, rates reach a level where refinancing begins to wane.
Suppose in a certain community, there will be M (r ) thousand refinanced mortgages when the
30-year fixed mortgage rate is r % , where
1  0.05r
M (r ) 
for 1  r  8
1  0.004r 2
a. For what values of r is M (r ) increasing?
b. For what interest rate r is the number of refinanced mortgages maximized? What is this
maximum number?
Solution (e.g. 3.1.7)
Differentiating M (r ) 
M (r ) 
1  0.05r
by the quotient rule, we get
1  0.004r 2
0.05(1  0.004r 2 )  (1  0.05r )0.008r 0.05  0.008r  0.0002r 2

(1  0.004r 2 ) 2
(1  0.004r 2 ) 2
By setting the numerator in this expression for M (r ) equal to 0, we find that r  5.495 is the
only solution of M (r )  0 in the interval 1  r  8 , and hence is the only critical number of M (r )
in its domain. The critical number divides the domain 1  r  8 into two intervals, 1  r  5.495
and 5.495  r  8 . Evaluating M (r ) at the test numbers in each interval (say, at r  2 and r  6 ),
we obtain the arrow diagram shown here.
+++++
1
-------5.495
8
u
The arrow pattern indicates that M (r ) is increasing when 1  r  5.495 and it increases to a
maximum r at r  5.495 , which means that the interest rate is 5.495%, and the maximum number
is M (5.495) 
1  0.05(5.495)
 1137 mortgages.
1  0.004(5.495) 2
72. Find constants a, b and c so that the graph of the function f ( x)  ax 2  bx  c has a relative
maximum at (5, 12) and crosses the y axis at (0, 3).
Solution
Since the function f(x) crosses through the points (5, 12) and (0, 3), we can get:
a(5) 2  b(5)  c  12
and
a(0) 2  b(0)  c  3
Since (5, 12) is the relative maximum, we have f (5)  0 , that is
2a(5)  b  0
From these three equations, we can solve for a, b and c to get: a  
9
18
,b  ,c  3
25
5
§3.2 Concavity and Points of Inflection
P216 9, 12, 23, 31, 37, 53, 57, 61
In problems 9 and 12, determine where the graph of given function is concave upward and concave
downward. Find the coordinates of all inflection points.
1
9. g (t )  t 2 
t
Solution (e.g. 3.2.1, 3.2.2)
The function g (t ) is undefined for t  0 .
g (t )  2t 
1
t2
g (t )  2 
2 2(t 3  1)

t3
t3
The second derivative g (t ) does not exist at t  0 and g (t )  0 for t  1 . These numbers
divide the x axis into three intervals on which g (t ) does not change sign, namely t  0 , 0  t  1
and t  1 . Evaluating g (t ) at test numbers in each of these intervals (say t  1 ,
t  0.5, and t  2 ) to get
g ( 1)  0
Sign of
g (0.5)  0 g ( 2)  0
g (t )
++++++++
--------------
++++++++
x
0
1
Inflection number
Thus, the graph of g (t ) is concave upward for t  0 and for t  1 and concave downward
1
for 0  t  1. Since the concavity changes at t  1 and g (1)  12   0 , it follows that (1, 0) is an
1
inflection point of g (t ) .
12. g ( x)  3x 5  25x 4  11x  17
Solution (e.g. 3.2.1, 3.2.2)
The function g (x ) is defined for all numbers t .
g ( x)  15x 4  100 x 3  11
g ( x)  60 x 3  300 x 2  60 x 2 ( x  5)
The second derivative g (x ) is continuous for all x and g ( x )  0 for x  0 and x  5 . These
numbers divide the x axis into three intervals on which g (x ) does not change sign, namely x  0 ,
0  x  5 and x  5 . Evaluating g (x ) at test numbers in each of these intervals (say x  1 ,
x  1, and x  6 ) to get
g ( 1)  0
g (1)  0 g (6)  0
Sign of g (x)
---------------
--------------
++++++++
x
0
5
Thus, the graph of g (x ) is concave downward for x  0 and 0  x  5 , and concave upward
for x  5 . Since the concavity changes at x  5 and g (5)  -6212, it follows that (5, -6212 ) is an
inflection point of g (x ) .
In Problem 23, determine where the given function is increasing and decreasing, and where its
graph is concave up and concave down. Find the relative extrema and inflection points, and sketch
the graph of the function.
23. g ( x)  x 2  1
Solution (e.g. 3.2.4)
The function g (x ) is defined for all numbers t .
g ( x) 
x
x2 1
1
g ( x) 
( x  1)
2
3
2
The only critical number is x  0 which divide the x axis into two intervals, on each of which the
sign of the derivative does not change; namely, x  0 and x  0 . Choose a test number c in each of
these intervals (say, -1 and 1 respectively) and evaluate g (c) in each case:
g (1)  0
g (1)  0
Thus, the graph of g (x ) falls for x  0 , and rises for x  0 , so there must be a relative minimum at
x  0 , and the relative minimum point is (0, 1).The second derivative g (x ) is continuous and
positive for all x . So g (x ) is concave upward for all real numbers x . Thus, there is no inflection
point.
Sign of g (x)
Sign of g (x)
-----------
++++++++
++++++++
x
x
0
Min
No inflection
y=g(x)
-1
0
1
x
In Problem 37 use the second derivative test to find the relative maxima and minima of the given
function.
( x  2) 3
37. f ( x) 
x2
Solution (e.g. 3.2.5)
The function f (x) is undefined for x  0 .
f ( x) 
( x  2) 2 ( x  4)
x3
f ( x) 
24( x  2)
x4
The first derivative f (x) is zero when x  2 and x  4 , the corresponding points (2,0) and (-4,
-13.5) are the critical points of f . Since f (4) 
24(4  2)
 0 , the critical point (-4, -13.5) is
(4) 4
a relative maximum, and since f (2)  0 , test fails for x=2 which is an inflection number.
53. MARGINAL AMALYSIS The cost of producing x units of a commodity per week is
C ( x)  0.3x 3  5x 2  28x  200
a. Find the marginal cost C (x) , and sketch its graph along with the graph of C (x) on the same
coordinate plane.
b. Find all value of x where C ( x )  0 . How are these levels of production related to the graph of
the marginal cost?
Solution (e.g. 2.5.1, 3.2.6)
The cost function is defined for all x  0 . The marginal cost is
C ( x)  0.9 x 2  10 x  28
which is positive for all x  0 . So there is no critical number of cost function. The second
derivative of the cost function is
C ( x)  1.8 x  10
C ( x )  0 when x  50 / 9 , which is the possible inflection number of cost function and is the
critical number of the marginal cost. The second derivative of the marginal cost is
C (3) ( x)  1.8
which is positive for all x  0 . So the marginal cost is concave upward for x  0 , and marginal cost
has a relative minimum at x  50 / 9 . All of these are indicated in the following diagrams.
Sign of C (x)
Sign of C (x )
C (x )
C (x )
++++++++
---------------
++++++++
x
0
x
0
50/9
Inflection number
Sign of C (x)
Sign of C (3) ( x)
C (x )
---------------
C (x )
++++++++
++++++++
x
0
50/9
0
x
Inflection number
Interpreting the diagrams, x=50/9 is the inflection number of the cost function, the graph of
marginal cost falls to the relative minimum at x=50/9.
400
C(x)
C’(x)
200
10
20
30
50/9
57. POPULATION GROWTH A 5-year projection of population trends suggests that t years from
now, the population of a certain community will be P(t )  t 3  9t 2  48t  50 thousand.
a. At what time during the 5-year period will the population be growing most rapidly?
b. At what time during the 5-yeat period will the population be growing least rapidly?
c. At what time is the rate of population growth changing most rapidly?
Solution (e.g. 3.2.6, 3.4.1)
The rate of growth is the derivative of the population P (t ) ; that is
R(t )  P(t )  3t 2  18t  48
a. Since the domain of the function P (t ) is all numbers 0  t  5 , the goal is to find the largest
rate R (t ) for 0  t  5 . The derivative of the rate function is
R (t )  6t  18
which is zero when t  3 , positive for 0  t  3 , and negative for 3  t  5 , as indicated in the
arrow diagram shown.
+++++
0
-------3
5
t
Thus, the rate R (t ) increases for 0  t  3 , decreases for 3  t  5 , and has the maximum value
when t  3 . That is the rate is largest when t  3 years.
b. Since R(0)  48 and R (5)  63 , so the rate is smallest when t  0 years.
c. The rate of population growth is P (t )  6t  18 , which is largest when t  0 .
61. HOUSING STARTS Suppose that in a certain community, there will be M (r ) thousand new
houses built when the 30-year fixed mortgage rate is r % , where
1  0.02r
M (r ) 
1  0.0 0 9r 2
a. Find M (r ) and M (r )
b. Sketch the graph of the construction function M (r ) .
c. At what rate of interest r is the rate of construction of new houses minimized?
Solution (e.g. 3.2.6)
a.
M (r ) 
0.02  0.018r  0.00018r 2
(1  0.009r 2 ) 2
M (r ) 
 0.018  0.00108r  0.00048r 2  0.00000324r 3
(1  0.009r 2 ) 3
where r  0 .
b.
M (r ) 
0.02  0.018r  0.00018r 2
18r 2  1800r  2000

(1  0.009r 2 ) 2
100000(1  0.009r 2 ) 2

2(9r 2  900r  1000)
(3r ) 2  300  3r  22500  23500


50000(1  0.009r 2 ) 2
50000(1  0.009r 2 ) 2

(3r  150) 2  23500
(3r  150  23500 )(3r  1 5 0 2 3 5 0) 0

2 2
50000(1  0.009r )
5 0 0 0(10 0.0 0 9r 2 ) 2
So M (r )  0 for r>0 only when r  50 
23500
23500
 1.0990 ( r  50 
 -101.0990
3
3
is less than zero, then discard it), which is the critical number of M (r ) . Using the property of
quadratic polynomial, we find that M (r )  0 for 0  r  1.0990 and M (r )  0 for r>1.0990. So
the graph of the function M (r ) is increasing for 0  r  1.0990 and decreasing for r>1.0990, and
there is a relative maximum when r  1.0990 . When M (r )  0 , r=7.10 (discard the roots which are
less than zero). Using the property of cubic polynomial, we find that M (r )  0 for 0  r  7.10
and M (r )  0 for r>7.10. So the graph of the function M (r ) is concave downward
for 0  r  7.10 and concave upward for r>7.10.
M
r
c. When M (r )  0 , the rate of construction of new houses is possible minimized. That is, r=7.10
(discard the roots which are less than zero). Using the property of cubic polynomial, we find that
M ( r )  0 for 0  r  7.10 and M ( r )  0 for r>7.10. So the graph of the function M (r ) falls
for 0  r  7.10 and rises for r>7.10. So there is a relative minimum of the rate of construction of
new houses when r  7.10 .
§3.3 Curve Sketching
P230 14, 16, 28, 55
In Problems 14 and 16, find all vertical and horizontal asymptotes of the graph of the given
function.
5x 2
14. g ( x)  2
x  3x  4
Solution (e.g. 3.3.1, 3.3.2)
g (x ) is a rational function, but its domain is all real numbers x , so there is no vertical asymptote of
the graph of g (x ) . Dividing each term in the rational function g (x ) by x 2 (the highest power of
x in the denominator), we find that
5x 2
5
 lim
5
2
x  x  3 x  4
x  1  3 / x  4 / x 2
lim g ( x)  lim
x 
and similarly,
5x 2
lim g ( x)  lim 2
5
x 
x  x  3 x  4
Thus, the graph of g (x ) has y  5 as a horizontal asymptote.
16. g (t ) 
t
t2  4
Solution (e.g. 3.3.1, 3.3.2)
g (t ) is undefined for t  2 and t  2 . For t  2 and t  2 we have
lim g (t )  lim
x 2
x 2
lim g (t )  lim
and
x 2
x 2
t
 
t 4
2
t
t2  4
 
So t  2 and t  2 are the vertical asymptotes of function g (t ) . Dividing each term in the function
g (t ) by t , we find that
lim g (t )  lim
t 
t 
t
t 4
2
1
 lim
t 
1 4 / t 2
1
and similarly,
lim g (t )  lim
t 
x
t
t2  4
1
Thus, the graph of g (t ) has y  1 as a horizontal asymptote.
In Problems 28, sketch the graph of the given function
28. f ( x) 
1
1 x2
Solution (e.g. 3.1.5, 3.3.3, 3.3.4)
Since
1
u
is defined only for u  0 , the domain of
will be the set of all t such
f
that 1  x 2  0 .Factoring the expression 1  x 2 , we get
1  x 2  (1  x)(1  x)
So f is defined for  1  x  1 . The only interpret is (0,1) . Since
lim  f ( x)  lim 
x 1
x 1
1
1 x2
 lim f ( x)  lim
x 1
x 1
1
1 x2
 
x  1 and x  1 are the vertical asymptotes of function. But since f is defined for  1  x  1 ,
there is no horizontal asymptote of the function. Derivative the function f to get
f ( x) 
x
(1  x 2 ) 3 / 2
f ( x) 
and
1  2x 2
(1  x 2 ) 5 / 2
1  x  1
The only critical point is (0, 1) and the intervals of increase and decrease and concavity are as show
Sign of f (x )
Sign of f (x)
++++++++
----------0
Min
-1
++++++++
1
++++++++
x
x
0
No inflection
-1
1
Interpreting these diagrams, we conclude that the graph of f (x) is concave upward for all
 1  x  1 but is falling for  1  x  0 while rising for 0  x  1 . Thus, the graph of f (x) has a
relative minimum at the origin (0, 1) and its shape there is a cup. The graph of f (x) is shown in
the following figure.
x=-1
-1
x=1
0
1
x
§3.4 Optimization
P230 3, 10, 19, 27, 41, 49
In problems 3 and 10, find the absolute maximum and absolute minimum (if any) of the given
function on the specified interval.
3. f ( x) 
1 3
x  9 x  2;0  x  2
3
Solution (e.g. 3.4.1)
From the derivative
f ( x)  x 2  9  ( x  3)( x  3)
we see that the critical numbers are x  3 and x  3 . But neither of x  3 nor x  3 lies in the
interval 0  x  2 . Compute f (x) at the endpoints x  0 and x  2 .
f ( 2)  
f ( 0)  2
34
3
Compare these values to conclude that the absolute maximum of f (x) on the interval 0  x  2
is f (0)  2 and the absolute minimum is f (2)  
10. g ( x) 
34
.
3
1
;0  x  2
x 9
2
Solution (e.g. 3.4.1)
From the derivative
g ( x)  
2x
( x  9) 2
2
we see that the critical numbers are x  0 . Compute g (x ) at the endpoints x  0 and x  2 .
g ( 0)  
1
9
g ( 2)  
1
5
Compare these values to conclude that the absolute maximum of g (x ) on the interval 0  x  2 is
1
1
g (0)   and the absolute minimum is g (2)   .
9
5
MAXIMUM PROFIT AND MINIMUM AVERAGE COST In Problem 27, you are given the
price p (q ) at which q units of a particular commodity can be sold and the total cost C (q ) of
producing the q units. In each case:
(a) Find the revenue function R (q ) , the profit function P (q ) , the marginal revenue R (q ) , and
marginal cost C (q ) . Sketch the graphs of P (q ) , R (q ) , and C (q ) on the same coordinate axes
and the level of production q where P (q ) is maximized.
(b) Find the average cost A(q)  C (q) / q and sketch the graphs of A(q ) , and the marginal cost
C (q ) on the same axes. Determine the level of production q at which A(q ) is minimized.
19. p(q)  180  2q; C (q)  q 3  5q  162
Solution (e.g. 3.4.4, 3.4.5, 3.5.6)
a. The revenue function is
R(q)  qp(q)  180q  2q 2
And the marginal revenue is R(q)  180  4q
The profit function is
P(q)  R(q)  C  (180q  2q 2 )  (q 3  5q  162)  q 3  2q 2  175q  162
The marginal cost C (q ) is
C (q)  3q 2  5
So the marginal profit is P(q)  R(q)  C (q)  3q 2  4q  175  (q  7)(3q  25) , which is zero
when q=7 (discard another root which is less than zero), and is positive when q<7 and negative
when q>7, as indicated in the arrow diagram shown.
Sign of
P(x )
P(x)
Sign of
+++++
0
--------
++++++++++
t
7
t
0
So the profit is maximized when q=7. P (q )  6q  5 is lese than zero for all q>0, so the graph
of profit function is concave downward for all q>0. Since
C (q)  6q  0
C ( 3) ( q )  6  0
we have
Sign of C (3) (q)
C ' ' (q )
Sign of
q>0
++++++++
++++++++++q
q
0
0
Then the graphs of P (q ) , R (q ) and C (q ) is shown in the following graph.
p
C (q )
P (q )
180
R (q )
3
7
(0,-162)
q
b. The average cost is A(q)  C (q) / q  q 2  3  162 / q for q>0. We find
A' (q)  2q  162 / q 2
which is zero for q>0 only when q=4.3267. Since
A' ' (q)  2  324 / q 3  0 when q>0
it follows that the graph of average cost A(q ) is concave upward for q>0. We have
A' (q )
Sign of
Sign of A' ' ( q )
-----------
++++++++
q
q
4.3267
0
++++++++++0
So A(q ) is minimized when q=4.3267. Then the graphs of A(q ) and C (q ) is shown in the
following graph.
p
A(q )
C (q )
3
4.3267
q
ELASTICITY OF EDMAND In Problem 27, compute the elasticity of demand for the given
demand function D ( p ) and determine whether the demand is elastic, or of unit elasticity at the
indicated price p .
27. D( p) 
3000
 100; p  10
p
Solution (e.g. 3.4.6)
The elasticity of demand for the demand function D ( p ) is
E ( p) 
Since | E (10) |
p dq
30

q dp p  30
30
3
  0 , the demand is elastic at price p  10 .
10  30 2
41. DEMAND FOR ART An art gallery offers 50 prints by a famous artist. If each print in the
limited edition is priced at p dollars, it is expected that q  500  2q prints will be sold.
a. What limitations are there on the possible range of the price p ?
b. Find the elasticity of demand. Determine the values of p for which the demand is elastic,
inelastic, and of unit elasticity.
c. Interpret the results of part (b) in terms of the behavior of the total revenue as a function of unit
price p .
d. If you were the owner of the gallery, what price would you charge for each print? Explain the
reasoning behind your decision.
Solution (e.g. 3.4.7)
a. Obviously q  500  2 p  0 . Since the offers 50 prints, we find that q  500  2 p  50 . So
0  500  2 p  50 , that is 225  p  250 .
b. The elasticity of demand is
E ( p) 
p dq
p
p

(2)  
q dp 500  2 p
250  p
and since 225  p  250 ,
 250   p  225
0  250  p  25
1
1

250  p 25
so
| E ( p) |
p
250
 1 
9
250  p
250  p
that is, the demand is elastic for 225  p  250 .
c. The demand is elastic for 225  p  250 , so the revenue is decreasing as price p increase. That is,
the specified percentage increase in price results in a larger percentage decrease in demand.
d. From the results of part (c), we find that the optimal price is $225, which is the endpoint of the
domain of the demand function. But if any number of prints are available, the p  125 maximizes
total revenue, since | E (125) | 1 .
49. BLOOD PRODUCION A useful model for the production p (x ) of blood cells involves a
function of the form
p ( x) 
Ax
B  xm
where x is the number of cells present, and A, B , and m are positive constants.
a. Find the rate of blood production R ( x)  p ( x) and determine where R ( x)  0 .
b. Find the rete at which R (x ) is changing with respect to x and determine where R ( x)  0 .
c. If m  0 , does the nonzero critical number you found in part (b) correspond to a relative
maximum or a relative minimum? Explain.
Solution
R( x)  p( x) 
a.
AB  A(1  m) x m
(B  x m )2
1/ m
 B 
which is zero when x  

 m 1
.
b. Derivative R (x ) to get
R' ( x) 
 Amx
m 1
(1  m) B 

Amx m1 (m  1)  x m 
[(1  m) x  (1  m) B]
m  1 


m 3
m 3
(B  x )
(B  x )
m
 B (m  1) 
which is zero when x  0 and x  
 m  1 
1/ m
 B (m  1) 
c. Use the first derivative test to classify the nonzero critical number x  
 m  1 
 B(m  1) 
divide the domain of R (x ) into two intervals. If 0  x1  
 m  1 
1/ m
 x2 ,
B (m  1)
m
 x2
m 1
(1  m) B
(1  m) B
m

 0  x2 
m 1
m 1
x1 
m
x1
so
m
Since A  0 and m  1, Am(m  1)  0 . Then we have
Amx1
m 1
 m (1  m) B 
 m (1  m) B 
m 1
(m  1)  x1 
Amx2 (m  1)  x2 

m 1 
m  1 



0

(B  x m )3
(B  x m )3
that is R' ( x1 )  0  R' ( x2 ) .
Sign of R(x)
-------0
++++++
++
 B(m  1) 
 m 1 


1/ m
x
1/ m
, which
 B (m  1) 
Interpreting the diagram, the graph falls to the relative minimum at x  
 m  1 
1/ m
.