Download Notation: • Cost in producing x items: C(x) • Average Cost (per item

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(x + h) − C(x)
0
C (x) = lim
h→0
h
C(x + h) − C(x)
≈
h
when h is small. If x is very large, then relatively speaking h = 1
would be small compared to x and we have
C(x + 1) − C(x)
= C(x + 1) − C(x)
C 0(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
Then
C(x) = 0.000002x3 − 0.03x2 + 400x + 80000.
• 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:
300000
1.8e+06
2e+06
1.6e+06
200000
1.4e+06
1.5e+06
100000
1.2e+06
x
1000
1e+06
1e+06
0
800000
–100000
600000
–200000
400000
500000
200000
–300000
0
0
1000
2000
3000
4000
x
5000
6000
7000
1000
2000
3000
4000
x
5000
6000
7000
2000
3000
4000
5000
6000
7000
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
pf 0(p)
E(p) = −
f (p)
Make sense? hmmm....
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”)
100 ∆x
% change in number demanded
x
=
% change in price
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.
pf 0(p)
E(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
3
s(t) = t2.
2
Velocity is the rate of change of position given by the derivative of
position:
ds
0
s (t) =
= 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
d2 s
s (t) = 2 = 3
dt
In this case acceleration is constant. We will later discuss how to
interpret the derivative in greater detail.
00
Notation is a bit confusing. Recall that “ dtd ” is shorthand for “find
are
the derivative with respect to t of what comes next.” dtd and dy
dt
not really 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”).
ds
d
s0(t) =
= (s(t))
dt dt
and
d2 s
d d
00
s (t) = 2 = ( s(t) )
dt
dt dt
A problem:
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:
√
3
Estimate 7.8 without using a calculator or any other computational device.
A problem, with special rules just for this example:
√
3
Estimate 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
1
0
p
f (x) =
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)?