Section 10.8 Marginal Analysis
Recall: the cost function C(x) gives the cost
in dollars to produce x items. C’(x) is the
marginal cost function, the rate of change of
the total cost or the cost of producing the
next item.
Example 1:
The cost of producing x teddy bears a day at
the Cuddly Companion Company is
calculated by their marketing staff to be
given by the function
C(x) = 100 + 40x – 0.001x2
a) Find the marginal cost function and use it
to estimate how quickly the cost is going up
at a production level of 100 teddy bears.
Compare this with the exact cost of
producing the 101st teddy bear.
We can find C’(100) with nDeriv(.
C’(100) = 39.80
Interpretation: The total cost of teddy bear
production is increasing at rate of $39.80 per
bear when 100 bears are produced or the
cost of producing the 101st bear is $39.80.
We can find the exact cost of producing the
101st bear using C(x):
Cost
Cost
for the first
–
for the first
101 bears
100 bears
C(101)
–
39.799
C(100)
b) Find the average cost function C (x ) and
evaluate C (100) . What does the answer tell
you?
Average cost of producing x items is
C x 
total cost of production

x
number of items produced
100
C ( x) 
 40  0.001x
x
C 100 
C (100) 
 40.9
100
This tells us that the average cost per bear
for the first 100 bears is $40.90.
Since the marginal cost is less than the
average cost, the cost of producing each
Teddy bear is decreasing when 100 bears
are made or the total cost is increasing more
slowly at a production level of 100 bears
than at lower production levels.
You can graph this.
0 ≤ X ≤ 40,000 and 0 ≤ Y ≤ 410,000
Example 2:
Assume that the demand function for tuna
in a small coastal town is given by
60
p  0.5
q
(200 ≤ q ≤ 800)
where p is the price of tuna and q is the
number of pounds of tuna that can be sold
at the price p in one month.
a) Calculate the price the town’s fishery
should charge for tuna in order to produce
a demand of 400 lbs of tuna per mo.
We need to find p when q equals 400.
60
p400  
3
0.5
400
The fishery should charge $3 per pound to
produce a demand of 400 pounds.
b) Calculate the monthly revenue as a
function of the number of pounds of tuna q.
Revenue = price x quantity
Rq   pq
60
 0. 5 q
q
 60q
0.5
c) Calculate the revenue and marginal
revenue at a demand of 400 pounds per
month and interpret the results.
Revenue:
0.5




R 400  60 400
Marginal revenue:
R' 400  30400
0.5
The revenue is $1,200 and increasing at a
rate of $1.50 per pound when production is
400 pounds.
d) If the town’s fishery’s monthly catch of
tuna amounts to 400 pounds, and the price
is at least the level in part a), would you
recommend that the fishery raise or lower
the price of tuna in order to increase its
revenue?
Since revenue here is increasing, and 400
pounds is constant, they can raise the price
and increase the revenue.