Download Solutions to Assignment 8

Solutions to Assignment 8
12.4, 2. The graph of a typical average cost function A(x) = C(x)/x, where
C(x) is a total cost function associated with the manufacture of x
units of a certain commodity is shown in the following figure (see page
733 of text).
a Explain in economic terms why A(x) is large if x is small and why
A(x) is large if x is large.
Solution: A(x) is large for small x since the fixed costs of production are spread out only over a few items. For x large, on the
other hand, the fixed costs of production are spread out over a
large number of units, making A(x)/x smaller.
b What is the significance of the numbers x0 and y0 , the x- and
y-coordinates of the lowest point on the graph of the function A?
Solution: The number x0 denotes the number of units of production that will produce the lowest average cost of production. The
number y0 denotes the lowest average cost of production possible.
12.4, 14 Pulsar also manufactures a series of 19-in. color television sets. The
quantity x of these sets demanded each week is related to the wholesale
unit price p by the equation
p = −0.00gx + 180
The weekly total cost incurred by Pulsar for producing x sets is
C(x) = 0.000002x3 − 0.02x2 + 120x + 60, 000
dollars. Answer the questions in Exercise 13 for these data.
a Find the revenue function R and the profit function P .
Solution: The revenue function is given by price multiplied by
quantity or p · x. Given the price function, this yields:
R(x) = −0.00gx2 + 180x
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This need not be simplified.
The profit function is found by subtracting the cost function C(x)
from the revenue function R(x) or
P (x) = R(x) − C(x)
= −0.00gx2 + 180x − (0.000002x3 − 0.02x2 + 120x + 60, 000)
= −0.000002x3 + 0.014x2 + 60x − 60000.
This need not be simplified.
b Find the marginal cost function C 0 , the marginal revenue function
R0 , and the marginal profit function P 0 .
Solution: Simply take the derivatives of C(x), R(x), and P (x)
to get
C 0 (x) = 60, 000 − 0.04x + 120
R0 (x) = −0.012x + 180
P 0 (x) = −0.000006x2 + .028x + 60.
c Compute C 0 (2000), R0 (2000) and P 0 (2000) and interpret your results.
Solution: From the above formulas, we have
C 0 (2000) = 64
R0 (2000) = 156
P 0 (2000) = 92.
The meanings are as follows: C 0 (2000) = 64 tells us that when
producing 2000 sets, the marginal cost of producing one more set
is $64.
R0 (2000) = 156 means that when producing 2000 sets, the marginal
revenue of producing another set is $156.
P 0 (2000) = 92 means that when producing 2000 sets, the marginal
profit from producing another set is $92.
d Sketch the graphs of the functions C, R, and P and interprets
parts (b) and (c) using the graphs obtained.
Solution: The three graphs are below, the red (highest until
x = 10000 is the revenue, middle is cost, lowest is profit)
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1.5 ´ 106
1.0 ´ 106
500 000
2000
4000
6000
8000
10 000
The answers in (b) give the slopes of each of these graphs as a
function of x. At x = 2000, the answers in (c) give the slopes at
x = 2000 these are telling us that each of the graphs the slope at
x-value 2000.
12.4, 30 The management of Titan Tire Company has determined that the
quantity demanded x of theiir Super Titan tires per week is related
to the unit price p by the equation
x=
q
144 − p
(0 ≤ p ≤ 144)
where p is measured in dollars and x in units of a thousand.
a Compute the elasticity of demand when p = 63, 96, and 108?
Solution: Recall that elasticity of demand is given by the formula
E(p) = −
pf 0 (p)
.
f (p)
To apply this formula, we need quantity x in terms of price p,
which we have been given. (However, in problem 33, this is not the
case!). We have that f 0 (p) = −(1/2)(144 − p)−1/2 . Consequently,
the elasticity of demand is:
p(−1/2)(144 − p)−1/2
√
144 − p
p
=
2(144 − p)
p
=
288 − 2p
E(p) = −
3
You should reduce the answer into one of the last two forms since
they are so much simpler. We now have
7
18
E(96) = 1
3
E(108) =
2
E(63) =
b Interpret the results obtained in part (a).
Solution: The result from part a tells you that:
When the price is $63 then if you raise the price by a given percent
y, the demand will decrease only 7y/18 percent, so that the net
revenue will increase. More generally stated, as the elasticity is
less than 1, the demand tends to decrease by a lower percent than
a small increase in price (the previous answer is better). Thus,
you can raise prices some and increase revenue.
When the price is $96 then if you raise the price by a given percent
y, the demand will decrease by the same percent, so that the
net revenue will stay stable. More generally stated, when the
elasticity is 1 the demand decreases by the same percent as the
price increases when the price is changed. (the previous answer
is better). When the price is $63 then if you raise the price by
a given percent y, the demand will decrease only 3y/2 percent,
so that the net revenue will decrease. More generally stated, as
the elasticity is greater than 1, the demand is overly responsive to
price increases (the previous answer is better). In this case, you
will gain more revenue if you decrease the price.
c Is the demand elastic, unitary, or inelastic when p = 63, 96, and
108?
Solution: The demand is inelastic at p = 63, unitary at p = 96,
and elastic at p = 108.
12.5, 10 Find the first and second derivatives of the function h(x) = (x2 +
1)2 (x − 1).
Solution: We can do this problem in two different ways. The first uses
the product rule, which gives
h0 (x) = 2(x2 + 1)2x(x − 1) + (x2 + 1)2
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=
=
=
00
h (x) =
2(x2 + 1)(2x2 − 2x) + x4 + 2x2 + 1
2(2x4 − x3 + 2x2 − x) + x4 + 2x2 + 1
5x4 − 4x3 + 6x2 − 4x + 1
20x3 − 121x2 + 12x − 4
You may leave the first in the form above, but I don’t recommend
trying to do the latter without simplifying.
The other method is to simplify h(x) at the beginning of the problem,
where you obtain that
h(x) = x5 − x4 + 2x3 − 2x2 + x − 1.
√
12.5, 26 Find the third derivative of the given function. g(t) = 2t + 3.
Solution: We write this as a power so g(t) = (2t + 3)1/2 . Then taking
the derivatives
1
(2t + 3)−1/2 2 = (2t + 3)−1/2
2
g 00 (t) = −(2t + 3)−3/2
g 000 (t) = 3(2t + 3)−5/2
g 0 (t) =
12.5, 34. The body mass index (BMI) measures body weight in relation to height.
A BMI of 25 to 29.9 is considered overweight, a BMI of 30 or more is
morbidly obese. The percent of the U.S. population that is obese is
approximated by the function
P (t) = 0.0004t3 + 0.00362 + 0.8t + 12
(0 ≤ t ≤ 13)
where t is measured in years, with t = 0 corresponding to the beginning
of 1991. Show that the rate of the rate of change of the percent of the
U.S. population that is deemed obese was positive from 1991 to 2004.
What does this mean?
Solution: The rate of the rate of change is the 2nd derivative, which
is
P 00 (t) = 0.0024t + 0.0072.
We note that P 00 (t) > 0 if 0.0024t + 0.0072 > 0. Clearly this is true
for all t ≥ 0, in which case, this is true from t = 0 to t = 13. This
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means that the percent of the U.S. population that is getting obese was
increasing more rapidly each year over the previous year for the entire
period. That is, the increase was greater each successive year between
1991 and 2004.
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