Notation:
• Cost in producing x items: C(x)
• Average Cost (per item): C̄(x) =
C(x)
x
• price: p . . . often related to x in a demand equation
• Revenue, x items: R(x) = xp
• Profit, x items: P (x) = R(x) − C(x)
The derivative of cost C is called the marginal cost.
Similarly, the derivative of C̄, R, or P are called the marginal average cost,
marginal revenue, and marginal profit respectively.
Why?
*******************
In common language, we associate the word “margin” with the idea of “just one
more.” (!?) This relationship is approximately correct as defined here using the
derivative. Recall that
C 0 (x) =
≈
C(x + h) − C(x)
h
C(x + h) − C(x)
h
lim
h→0
when h is small. If x is very large, then relatively speaking h = 1 would be
small compared to x and we have
C 0 (x) ≈
C(x + 1) − C(x)
= C(x + 1) − C(x)
1
so the derivative is approximately the same as the cost of “one more item” (i.e.
the difference between producing x items and x + 1 items).
Suppose that the cost of producing x LED televisions is given by
C(x) = 0.000002x3 − 0.03x2 + 400x + 80000.
Then
• C 0 (x) =?
• C 0 (2000) =?
• C(2000) =?
• C(2001) =?
*******************
Suppose that the cost of producing x LED televisions is given by
C(x) = 0.000002x3 − 0.03x2 + 400x + 80000.
The demand relationship between price p and the number x we estimate that
we can sell at that price is
p = 600 − 0.05x
• Find formulas for revenue and profit.
• Find marginal revenue and marginal profit.
• Evaluate R0 (2000) and P 0 (2000)
**********************
Graphs of cost, revenue, and profit:
**********************
Suppose that we write the demand equation relating price p and the number we
can produce and sell x in the form
x = f(p)
Then the elasticity of demand E is defined to be
E(p) = −
Make sense? hmmm....
**********************
pf 0 (p)
f(p)
Let ∆p be a change in price and let ∆x = f(p + ∆p) − f(p) = “demand at new
price - demand at old price” (where “demand” = x = “number we can sell”)
% change in number demanded
% change in price
=
=
=
≈
≈
100 ∆x
x
100 ∆p
p
f(p+∆p)−f(p)
f(p)
∆p
p
f(p+∆p)−f(p)
∆p
f(p)
p
0
f (p)
f(p)
p
0
pf (p)
f(p)
**********************
We expect the quantity on the previous page to be negative (why?), so put a “
- ” in front to make it positive.
E(p) =
≈
pf 0 (p)
f(p)
% change in number demanded
−
% change in price
−
• E(p) ≈ 0 means demand is not very sensitive to changes in price - Inelastic demand.
• E(p) large means demand is very sensitive to changes in price - Elastic
demand.
• E(p) = 1 means demand is unitary
**********************
The equation relating demand and price (price is in hundreds of dollars) for
exercise bicycles is
√
p = 9 − 0.02x,
for 0 ≤ x ≤ 450
For what price or range of prices is demand unitary? elastic? inelastic? (note
that 0 ≤ x ≤ 450 corresponds to ? ≤ p ≤?)
**********************
Recall: We imagine a bicycle whose position at time t was given by
s(t) =
3 2
t .
2
Velocity is the rate of change of position given by the derivative of position:
s0 (t) =
ds
= 3t
dt
Acceleration is the rate of change of velocity given by the derivative of the
velocity, which is actually the second derivative of position
s00 (t) =
d2 s
=3
dt2
In this case acceleration is constant. We will later discuss how to interpret the
derivative in greater detail.
**********************
d
Notation is a bit confusing. Recall that “ dt
” is shorthand for “find the
d
and dy
are not really
derivative with respect to t of what comes next.” dt
dt
fractions, and we are certainly not multiplying, but you can at least work the
notation as if you were. To the extent that we imagine that “dt” has any
meaning at all, we do think of it as a single unit (not a separate “d and t”).
s0 (t) =
ds
d
= (s(t))
dt
dt
and
d2 s
d d
= ( s(t) )
2
dt
dt dt
******************* A problem:
s00 (t) =
Water is being poured into a conical reservoir at the rate of 10 cubic feet per
minute. The reservoir has a diameter of 24 feet across the top and a depth of
4 feet. At what rate is the depth of the water increasing when the depth is 2
feet?
Something to keep in mind: The volume, radius, and depth of the water are all
functions of time.
**********************
A problem, with special rules just for this example:
Estimate
√
3
7.8 without using a calculator or any other computational device.
√
2
dy
If y = 3 x, then dx
= 13 x− 3 and the tangent line at the point (8, 2) has slope =
?
Write the equation for this line.
*******************
A more realistic problem using the ideas of the previous problem (calculators
OK):
Suppose that we do not have a formula for function f, but we do know that
f 0 (x) =
1
p
1 + f(x)
and that f(0) = 5. Can you estimate the value of f(0.2)?
Using your estimate for f(0.2), what could you do to estimate f(0.4)? From
there, how would you proceed if what we really wanted was f(1)?