ECON30020 Mathematical Economics Tutorial 6. Derivatives of Functions of One Variable
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ECON30020 Mathematical Economics
Tutorial 6. Derivatives of Functions of One Variable
Question 1. Average Costs and Marginal Costs
Denoting by C(x) the cost of producing the quantity x, one can define the following:
C(x)
.
2. The marginal cost function as MC(x) = C\(x).
Now, prove the following theorem:
Theorem. Suppose that the cost function C(x) has a continuous first derivative . Then the following is true:
(a) if MC > AC then AC is increasing.
(b) if MC < AC then AC is decreasing.
(c) at any interior minimum point x0 of AC,
AC(x0 ) = MC(x0 ).
Illustrate part (c) of the theorem for the following cost function:
C(x) = x3 - 20x2 + 120x.
Then discuss, lightheartedly, the following statement: “As Paul Samuelson changed from physics to economics, the average IQ in both disciplines increased.”
Answer: To prove the three points in the theorem, we first need to obtain the derivative of the average cost function using the Quotient Rule:
AC\(x) = ╱ 、 = = = .
(a) Suppose that MC(x) > AC(x). Then MC(x) - AC(x) > 0 and AC\(x) > 0; thus, average
cost AC(x) is increasing for values of x where MC(x) is greater than AC(x).
(b) Suppose that MC(x) < AC(x). Then MC(x) - AC(x) < 0 =→ AC\(x) < 0 =→ AC is
decreasing.
(c) Let x0 be an interior minimum point of the average cost function (supposing that one exists).1 Then at that value of x0 , the first-order condition must be satisfied: AC\(x0 ) = 0. From this,
AC\(x0 ) = 0 = MC(x0 ) - AC(x0 )
=→ AC(x0 ) = MC(x0 )
as required.
Note that this theorem proves the well-known result from Intro and Interemediate Microeco- nomics: if AC curve is U-shaped, then MC curve cuts it from below at the point where AC function achieves its minimum:
The relationship between marginal and
average costs
• MC < AC → AC decreases
the cost of the last unit is smaller than the cost of an average unit
• MC > AC → AC increases
the cost of the last unit is larger than the cost of an average unit
For U-shaped AC: the marginal cost passes through the minimum point of AC. |
q
We can illustrate part (c) for the following cost function: C(x) = x3 - 20x2 + 120x. The corresponding average cost is AC(x) = x2 - 20x + 120.
The FOC gives us its minimum: AC\(x) = 2x - 20 = 0 =→ x* = 10. (In order to verify that this is indeed a minimizer, we can check the second-order condition: AC\\(x) = 2 > 0, which is satisfied.) Therefore, average cost function has a minimum at x* = 10. The marginal cost is given by MC(x) = C\(x) = 3x2 - 40x + 120. Evaluating it at x* = 10 gives us MC(x = 10) = C\(10) = 3(10)2 - 40(10) + 120 = 20, which is equal to AC(x = 10) = 20. Indeed, at the value of output where AC achieves its minimum, it is equal to the MC . As for the Paul Samuelson statement, the point of the joke is that if you take away a below average item from one group and add an above average item to another group, the average increases in both. The underlying assumption is that Paul Samuelson has a lower IQ than the average physicist, but a higher IQ than the average economist, implying that physicists are smarter than economists.
Question 2. Continuity and Differentiability
Consider the function f : R t R, defined as
f (x) =.,x2 , x < 3
(x3 - 21x + 45, x > 3
(a) Is it continuous at x = 3? Use definition (4.3) to determine.
(b) Is it differentiable at x = 3? Use definition (5.5) to determine.
Note that you do not need to use ε - δ definition to find limits; instead, please use the fact that polynomials are continuous functions.
You might find the following formulae useful: (a + b)3 = a3 + 3a2 b + 3ab2 + b3 and (a + b)2 = a2 + 2ab + b2 .
Answer:
(a) Continuity. The conditions for the continuity of the function f (x) at point z are (1) the
limit of the function at point z must exist and (2) it must be equal to the value of the function at that point.
The condition (1) itself requires left-hand limit (LHL) to exist, right-hand limit (RHL) to exist, and for them to be equal to each other.
Therefore, we are checking that LHL = RHL = f (3), or, in proper math notation,
lim f (x) = lim f (x) = f (3).
z→3兰 z→3+
We have (using the fact that polynomials are continuous functions):
f (3) = 32 = 9,
lim f (x) = lim x2 = 32 = 9,
z→3兰 z→3兰
lim f (x) = lim (x3 - 21x + 45) = 33 - 21 * 3 + 45 = 9. z→3+ z→3+
As both limits exist and are equal to each other and to the value of the function at x = 3, the function is continuous at this point.
(b) Differentiability. We need to calculate two limits and show that they are equal to each other: LHL = the slope on the left f/-(z) and RHL = the slope on the right f(z), at x = 3.
Note that f (3) = 32 = 9.
Using the fact that polynomials are continuous functions, we obtain
f/-(3) = lim f (3 + △) - f (3) = lim (3 + △)2 - 32 = lim 32 + 6△ + △2 - 32 =
= lim (6 + △) = 6.
△→0兰
f(3) =△(l)+ =△(l)+ =
= lim = lim =
= lim (6 + 9△ + △2) = 6.
△→0+
As both limits exist and are equal to each other, the function is differentiable at x = 3 and the derivative at this point is f\ (3) = 6.
Question 3. The natural logarithm and some of its properties
Consider the function
y = f (x) =^ln(x).
where ln(.) denotes the natural logarithm.
(a) Compute the derivative of f (x) using the Chain Rule.
(b) For what values of x is the function defined? Show your working.
(c) Use L’Hˆopital’s rule (discovered by Johann Bernoulli, by the way) to find the following limit:
lim x ln(x).
z→0+
Answer:
(a) The derivative of y = ^ln x is
dy 1 d(ln x) 1 1 1
= × = × =
dx 2^lnx dx 2^lnx x 2x^lnx .
Here are more details on using the Chain Rule (if your require some). Let u = ln x. Then the Chain Rule states that
dy dy du
=
dx du dx ,
where in this case we have y = ^u and so
dy = 1 u-1/2 and du = 1
Hence,
dy 1 1 1
= × =
dx 2^lnx x 2x^lnx .
(b) First, observe that the square root function ^u is defined for values of u > 0. Then, the
function ln x is defined for values of x > 0. Since u = ln x, this is equivalent to saying that we want values of x such that
ln x > 0 ←→ eln z > e0 ←→ x > 1.
That is, the function y = ^ln(x) is defined for x > 1.
(c) When x approaches zero, function f (x) = x goes to zero and the function g(x) = ln(x) goes to minus infinity. What would happen to the product of the two? Which function is “more powerful”? Which one wins?
Rearranging and replacing the functions with their derivatives, we obtain:
z x ln(x) = z+ 1
x
= lim z→0+ ╱ 、
= lim z→0+
= z(-x) = 0.
For small values of x, the function f (x) = x is “more powerful” than ln(x), and “kills” it.
Question 4. Elasticities
The price elasticity of a demand function q = D(p), denoted ε(p), is defined as
dq/q dq p D/(p)p
dp/p dp q D(p)
and has the interpretation of giving the percentage change in quantity demanded when there is a one percent increase in price.
Find the expression for the price elasticity of demand ε for the demand function D(p) = 200 - 5p.
Determine the ranges of price for which the absolute value of ε is less than 1 and greater than 1. Illustrate on a graph of this demand function.
Answer: For the demand function q = D(p) = 200 - 5p, we have D/(p) = -5. Thus, the price elasticity of demand ε is equal to
dq p D/(p)p 5p p
dp q D(p) 200 - 5p 40 - p .
Note that it is negative when the demand curve is downward sloping. Thus, if the absolute value of elasticity is greater than / equal to / less than one, we have
|ε| ≥ 1 ←→ ≥ 1 ←→ p ≥ 40 - p ←→ p ≥ 20.
Therefore:
❼ |ε| < 1 for 0 < p < 20; inelastic demand
❼ |e| = 1 for p = 20
❼ |e| > 1 for 20 < p < 40; elastic demand
These results are presented on the graph below.
2022-06-18