9e.4.3.Pr61. The cost C(x) of producing the first x units of a
commodity is C(x) = e0.2 x.
a.
Find the marginal cost [dC/dx].
*
0.2 * e0.2 x
b.
Determine the level of production x for which the
marginal cost equals average cost A(x) = [C(x)/x] = [dC/dx].
*
Solve e0.2 x/x = .02 e0.2 x. x = 5. Average cost is minimized!
c.
Determine the level of production x for which the
average cost A(x) is minimized.
*
[dA/dx] = [(e0.2 x(.2x − 1))/(x2)], critical number at x=5.
d.
Here is the graph of C(x) = e0.2 x (I have
changed x to q).
For various values of q, use a straight edge or ruler to
represent lines through O = (0,0) and (q,C(q)). The slope of
these lines is C(q)/q = A(q). As you move q to the right, the
slope of the line from 0 to (q, C(q)) decreases and then
increases. Which of these lines has minimal slope? How does your
answer correspond the the result in parts (b.) and (c.)?
Here is an animated picture:
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2.
9e.4.2.Pr55.Variation. Worker Efficiency. Worker efficiency is
measured by (hourly) output Q and is given by a formula of the
form
Q(t)
= 500 − A e−kt,
where t is the "experience" measured by the time in months
after initial training.
a.
What is limt → ∞ Q(t)?
*
Since limt → ∞ e− k t = 0, limt → ∞ Q(t) = 500.
b.
It is observed that a newly trained worker has an
hourly output of 300 units, and that after 6 months, a worker
has an hourly output of 410 units. Find the function of the form
Q(t)
= 500 − A e−kt,
which fits the data.
Hint: What is the form of the function F(t) = 500 − Q(t)?
*
F(t) = 500 − Q(t) = A e−kt is an exponential decay function. Since F(0) = 500 − 300 = 200, and F(6) = 90,
F(t) = 200 * (90/200)t/6 . Q(t) = 500 − 200 * (90/200)t/6. Since (90/200)1/6 = eln(90/200)1/6 = e(1/6)ln.45 ≈ e−.13,
an alternate form is Q(t) = 500 − 200 e− .13 t .
3.
9e Example 4.3.14 Variation. Total net profit comes from wheat,
steel, and oil.
At a certain time,
The profit W from wheat accounts for 20% of net profit and is
increasing at a rate of 2%.
The profit S from steel accounts for 30% of net profit and is
increasing at a rate of 3%.
The profit O from oil accounts for 50% of net profit and is
decreasing at a rate of 5%.
At what percentage rate is total net profit increasing
or decreasing?
Hint: When the data is taken, find
a.
[W/(W + S + O)] =
*
.20 from the
data
b.
[([dW/dt])/W] =
*
.02
*
Similarly for S and O. You are asked to find
dW
dt
+
dS
dt
+
dO
dt
W + S + O
=
dW
dt
W
·
W
W + S + O
+ …
= (.20*.02 + .30*.03 − .50*.05)
= − .012 = −1.2 %
Data and results are given as proportions.
4.
9e.4.3.Pr66. Money is deposited in a bank offering interest
at an annual rate of 6% compounded continuously (CC). Find the
percentage rate of change with respect to time.
*
P(t) = P(0) e.06 t, [([dP/dt])/P] = [(.06 P(0) e.06 t)/(P(0) e.06t)] = .06 = 6%. Notice that P(0) e.06 tcancels. The
percentage rate (proportion) does not depend on the initial
investment.
5.
9e.4.3.Pr33. If g(u) = ln(u + √{u2 + 1}), find
g′(u). Simplify your answer.
