SOLUTION MANUAL

Introductory Mathematical Economics 2nd EditionB. Wade Hands

Mathematical Economics Tutorial StudentsLake Forest College

Spring 2006

Last Updated: February 1, 2006

CHAPTER ONE

1.1 Q = BP–1 with B > 0. This can be rewritten as P = BQ–1.

(a) TR(Q) = PQ = BQ–1Q = B. Thus, MR(Q) = ∂TR / ∂Q = 0.

(b) Notice that demand is undefined when Q = 0. Thus, MR = 0 only applies for Q > 0.

1.2 Consider Q = aP + b where a, b > 0. Solving for P we have .

(a) , so .

(b) Thus, graphically, the marginal revenue curve, which is twice as steep as the (inverse) demand

function, is:

Quantity (Q)

Price (P)

-b / a

MR = (2Q – b) / a(slope = 2 / a)

P = (Q – b) / a(slope = 1 / a)b / a

b

1.3 Constant elasticity demand curve: Q = APa with a < 0 and A > 0. Thus, P = (Q/A)1/a.

(a) , so .

(b) Notice that (Q/A)1/a > 0, so the sign of MR depends on the sign of (1+a) / a. It follows immediately

that MR = 0 if a = –1, MR > 0 if a < –1, and MR < 0 if –1 < a < 0.

1.4 Consider a supply curve Q = aP + b where a > 0 so that the supply curve is upward sloping. The inverse

supply curve is then P = (Q – b) / a. Thus, the supply curve cuts the price (vertical) axis if –b / a > 0 which

requires that b < 0. In contrast, the supply curve cuts the quantity (horizontal) axis if b > 0. Now,

.

It immediately follows that supply is elastic (that is, ES > 1) if b < 0 so that –b / Q > 0; that supply is

inelastic (that is, ES < 1) if b > 0 so that –b / Q < 0; and that supply is unit elastic (that is, ES = 1) if b = 0 so

that –b / Q = 0.

1.5 Demand for real money balances, m is m = e–απ where π is the expected rate of inflation and α > 0.

(a) .

(b) Suppose the demand for real money balances is unit elastic. As there is a negative relationship

between money demand and inflation (i.e., consumers want to hold less money when inflation is

expected to increase), this supposition is the same as supposing that εm,π = –1. In this case, the

supposition reduces to assuming that –1 = –απ, or that π = 1/α.

1.6 Given: q1 = f(P), q2 = g(P), and Q = q1 + q2 = f(P) + g(P). so now,

, , and

.

2

Thus,

.

3

1.7 Given: x* = M / 2px =0.5M(px)–1. Thus,

1.8 (a) TC(y) = y3 – 60y2 + 1210y + 600. Thus, MC(y) = ∂TC(y) / ∂y = 3y2 – 120y + 1210 and AVC(y) =

VC(y) / y = y2 – 60y + 1210. To find the minimums/maximums, consider the derivatives of MC and

AVC. MC΄ = 6y – 120. Set equal to zero, this implies that y = 20. Likewise, AVC΄ = 2y – 60. Set

equal to zero, this implies that y = 30. Further, as the coefficient on y2 is positive for both equations,

both are U-shaped parabolas (and not inverted U-shaped parabolas). Graphically, we have:

(b) TC(y) = y3 / 3 – 50y2 + 1500y + 50. Thus, MC(y) = ∂TC(y) / ∂y = y2 – 100y + 1500 and AVC(y) =

VC(y) / y = y2 / 3 – 50y + 1500. To find the minimums/maximums, consider the derivatives of MC and

AVC. MC΄ = 2y – 100. Set equal to zero, this implies that y = 50. Likewise, AVC΄ = 2y/3 – 50. Set

equal to zero, this implies that y = 75. Further, as the coefficient on y2 is positive for both equations,

both are U-shaped parabolas (and not inverted U-shaped parabolas). Graphically, we have:

Quantity (y)

$MC

1210AVC

20 30

Quantity (y)

$MC

3500AVC

50 75

4

1.9 Suppose TC(y) = ay3 + by2 + cy + d where a, b, c, and d are scalars. We are interested in finding the

necessary restrictions on a, b, c, and d so that MC and AVC are U-shaped and strictly positive for y > 0.

First, MC = 3ay2 + 2by + c and AVC = ay2 + by + c. Next, for both curves to be U-shaped, clearly a > 0 is

required. Otherwise the curves would be inverted U-shaped.

Second, in order for the curves to be U-shaped when y > 0, we need both curves to take a minimum when

y > 0. To insure this, set the derivative of both curves equal to zero; solve for y; and require this y to be

positive. In particular, MC΄ = 6ay + 2b = 0 requires y = –b / 3a. Requiring this to be positive, therefore,

requires b < 0 as we already have that a > 0. Likewise, AVC΄ = 2ay + b = 0 requires y = –b / 2a. Requiring

this to be positive, therefore, also requires b < 0 as we already have that a > 0.

Third, in order to insure that all values are positive when y > 0, it is enough to make sure that the values of

the functions are positive at these minimums. For MC, this requires that 3a(–b/3a)2 + 2b(–b/3a) + c > 0,

which requires c > b2 / 3a. For AVC, this requires that a(–b/2a)2 + b(–b/2a) + c > 0, which requires

c > b2 / 4a. As the first requirement is stronger, we need that c > b2 / 3a.

In summary, the requirements on the parameters are that a > 0, b < 0, and c > b2 / 3a. There are no

restrictions on d.

1.10 Let y = 15L2 – L3.

(a) MPL = dy / dL = 30L – 3L2.

(b) Graphically, this is an inverted U-shaped parabola:

Labor (L)

Output (y)

75

MPL

5 10

5

1.11 Equation (1.36) gave us C(Y) = mYd + C0, and solving the model gave us

where A = C0 + I0 + G0.

(a) The autonomous tax multiplier =

(b) The balanced budget multiplier =

1.12 Mark-up = m = (P – MC) / P.

Recall that MR(Q) = P + (∂P/∂Q)∙Q and that ε = –(∂Q/∂P)∙(P/Q). Thus, MR = P∙( 1–1/ε). Moreover, the

profit maximizing condition is that MR = MC. Thus, we have that

which implies that , or, that .

1.13 We have that TC(Q) = 200Q – 24Q2 + Q3. Thus, MC(Q) = 200 – 48Q + 3Q2. Under perfect competition,

we know that P = MC. Thus, P = 200 – 48Q + 3Q2. Further, in the long-run under perfect competition,

profit equals zero. So now we have:

Π = PQ – TC(Q)

Π = (200 – 48Q + 3Q2)Q – [200Q – 24Q2 + Q3]

Π = 200Q – 48Q2 + 3Q3 – 200Q + 24Q2 – Q3

Π = 2Q3 – 24Q2 = 0

Thus, Q = 0 or Q solves 2Q – 24 = 0, which gives us Q = 12. At this quantity, P = MC = 200 – 48(12) +

3(12)2 = $56.

1.14 For each of the following demand and cost curves, we are to find the profit maximizing quantity. For each,

we will find MR and MC, set these equal, and solve for Q*.

(a) P = 10 – 0.1Q implies that MR = 10 – 0.2Q.

TC(Q) = 2Q + 0.025Q2 implies that MC = 2 + 0.05Q.

Setting MC = MR and solving for Q yields Q* = 32,

6

(b) P = 22 – 2Q implies that MR = 22 – 4Q.

TC(Q) = Q3/3 – 10Q2 + 50Q + 45 implies that MC = Q2 – 20Q + 50.

Setting MC = MR and solving for Q yields Q2 – 16Q + 28 = 0. This factors to (Q–14)(Q–2) = 0. It is

easy to show that Q = 2 minimizes profits whereas Q = 14 maximizes profits. Therefore, Q* = 14.

(c) Q = –P + 25 implies that P = 25 – Q, which implies that MR = 25 – 2Q.

TC(Q) = Q3/12 – 2.5Q2 + 30Q + 100 implies that MC = 0.25Q2 – 5Q + 30.

Setting MC = MR and solving for Q yields Q2 – 12Q + 20 = 0. This factors to (Q–10)(Q–2) = 0. It is

easy to show that Q = 2 minimizes profits whereas Q = 10 maximizes profits. Therefore, Q* = 10.

(d) Q + 2P – 90 = 0 implies that P = 45 – 0.5Q, which implies that MR = 45 – Q.

ATC(Q) = Q2/2 – 20Q + 82.5 + 125/Q implies that VC = Q3/2 – 20Q2 + 82.5Q which implies that

MC = 1.5Q2 – 40Q + 82.5.

Setting MC = MR and solving for Q yields Q2 – 26Q + 25 = 0. This factors to (Q–25)(Q–1) = 0. It is

easy to show that Q = 1 minimizes profits whereas Q = 25 maximizes profits. Therefore, Q* = 25.

1.15 Demand of P = –bQ + c implies that MR = c – 2bQ. Moreover, AVC = a implies that VC = aQ and that

MC = a. Thus, setting MR = MC, we have that c – 2bQ = a so that Q* = (c – a) / 2b. This holds as long as

c > a, which makes sense as c is the maximum willingness to pay and a is the marginal cost of production.

Finally, P* = –b(c – a) / 2b + c = (c + a) / 2.

1.16 Suppose Q = L1/2 so that L = Q2. The cost of labor is fixed at $4, and the firm has a fixed cost of $100.

(a) Total cost = TC(Q) = wL + 100 = 4L + 100 = 4Q2 + 100.

(b) Average total cost = TC(Q) / Q = 4Q + 100/Q.

(c) Marginal cost = ∂TC(Q) / ∂Q = 8Q.

(d) Let P = –8Q + 96 so that MR = 96 – 16Q. Then setting MR = MC, we have that 96 – 16Q = 8Q or that

24Q = 96 so that Q* = 4. At this quantity, we know that P* = $64.

7

1.17 Suppose y = –L2 + 10L. A perfectly competitive firm faces a constant price of $10 and a constant wage rate

of $40. Thus, π(L) = py – wL = 10(10L – L2) – 40L = 60L – 10L2. Therefore,

Solving this gives us the optimal amount of labor to employ is L* = 3.

1.18 The supply of labor is given by w = L1/2. Production is characterized by y = 2L1/2. And the price of the

output is constant at $13.50 per unit.

Thus, π(L) = py – wL = 13.50(2L1/2) – L1/2(L) = 27L1/2 – L3/2. Therefore,

Solving this gives us the optimal amount of labor to employ is L* =

9. At this level of employment, the firm faces a wage of 3.

1.19 Consider the standard consumer choice problem with U(x,y) = x + ln(y).

(a) The budget constraint is x∙px + y∙py = M. Thus, x = (M – y∙py) / px. Optimal choice further requires that

Thus, as Ux = 1 and Uy = 1/y, we have that . Thus, y* = px / py and

x* = (M – (px / py.)∙py) / px = (M – px) / px = (M/px) – 1.

(b) Given these optimal demand curves, we can find the standard six elasticities:

8

9

(c) so, if px = 0, then the own-price elasticity of the demand for good x would be unit

elastic.

1.20 There are two goods, x and y. There is a fixed amount of labor: Lx + Ly = 1. And the production functions

are such that x = 2Lx1/2 and y = 2Ly

1/2 so that Lx = x2/4 and Ly= y2/4. The production possibilities curve,

which is always graphed in terms of the two goods, not labor, is found by simple substitution. In particular,

Lx + Ly = 1 implies that . Thus, so that .

Now, to check for concavity, notice that and then

which is clearly negative. Thus, the production

possibilities curve is concave.

1.21 We have that C = mY + C0. Let 0 < θ < 1 and let y0 and y1 be real numbers. Now, consider

and notice that

Thus, C(y) is both concave and convex.

1.22 Let F be fixed costs. Then AFC = F / y. We aim to show that AFC is strictly convex in two different ways.

Second Derivative Test: AFC΄ = –F∙y–2 so that AFC΄΄ = 2F∙y–3 > 0, and thus AFC is strictly convex.

Definition: Consider any y0 and y1 both positive and y0 < y1. Suppose that AFC is not strictly convex. This

requires that there exist some y1 where y0 < y1 and AFC(y1) ≤ AFC(y0) + (y1–y0)AFC΄(y0). Thus, using that

AFC = F / y and AFC΄ = –F∙y-–2 we have that

,

which reduces to . This further reduces to as (y0–y1) < 0. But this equation

implies that y0 ≥ y1, which is a contradiction.

10

1.23 Suppose the wage is fixed at w̃ so that the real wage is w = (1 – t)w̃. We also know that the labor supply

curve, S(w) is upward sloping so that ∂S(w)/∂w > 0. Finally, the elasticity of labor supply is:

.

Now, as T = t∙S(w)∙w̃:

Finally, for the Laffer effect to hold, tax revenue must fall as the tax rate increases, i.e., ∂T/∂t < 0, it must

be that which requires .

1.24 (a) Von Thünen’s definition of the real rate of interest, r, is

which is the classical definition of the rate of interest being the return on advanced labor costs.

(b) For von Thünen, savings equals wage income less consumption, which says that workers do all of the

savings.

(c) Maxw r(w)∙S(w) = .

Thus, the first order condition is: –1 + paw–2 = 0, which solves as w* = (pa)1/2 as required.

1.25 Let w = r = 1 and y = f(L,K) = LK / 100.

(a) MinL,K L + K such that y = 0.01LK, which requires K = 100y / L.

MinL L + 100y / L.

The first order condition is 1 – 100y / L2 = 0. The solution is that L* = 10y1/2 and K* = 10y1/2.

11

(b) LRTC(y) = wL* + rK* = 10y1/2 + 10y1/2 = 20y1/2.

12

(c) LRATC(y) = LRTC(y) / y = 20y–1/2. LRMC(y) = ∂LRTC(y) / ∂y = 10y–1/2. Graphically, these curves

look like the following:

(d) Yes. The production function exhibits increasing returns to scale as y = LK (the exponents sum to

more than 1). Thus, as the firm employs more and more inputs, it makes output at an even faster rate

(which lowers the per-unit cost).

1.26 (a) Suppose (p1 + wt1)x1 + (p2 + wt2)x2 = wT and U(x1, x2) = x1x2. Let t2 = 0 and p2 = 1, then the budget line

requires that x2 = wT – (p1 + wt1)x1. The optimization problem then becomes:

Max x1 ∙ [wT – (p1 + wt1)x1]

The first-order condition is:

wT – 2p1x1 – 2wt1x1 = 0.

The solution is and .

(b) , which makes

sense. As one’s wage increases, one purchases more golf even though there is limited time.

Output (y)

Cost ($)

LRMCLRATC

13

CHAPTER TWO

2.1

2.2

2.3 Consider the utility function and the inequality from (2.26) to verify diminishing

marginal rate of substitution .

By computing it will be possible to directly verify the inequality.

Plugging straight into the inequality (2.26) it can be shown that

2.4 Consider the additively separable utility function From (2.26),

for the utility function U(x) to exhibit a diminishing marginal rate of substitution for any two goods. Now,

since U(x) is additively separable U12=0. So the above condition becomes

Diminishing marginal utility for all goods implies Therefore, since it follows

that the above condition will hold. Thus, diminishing marginal utility for all goods is sufficient for

diminishing marginal rates of substitution when the utility function is additively separable.

2.5

14

2.6 (a) Proof: First, recall

Then, we know Differentiating with respect to M, we get

`

(b) Proof: Recall , , and .

Differentiate the first equation with respect to ,

2.7 (a) Consider the function with , and .

To check for the degree of homogeneity the function needs to be considered.

In this case which implies the function y is homogeneous of

degree

15

(b) Consider the function with and ,

then ,

but no homogeneity can be detected.

(c ) Consider the function and

, then which implies the

function y is homogeneous of degree .

2.8

2.9

2.10

2.11 (a) Consider the Cobb-Douglas production function

From problem 2.10,

And recall that Using this fact, we calculate and

These yields:

and

From here we can calculate :

(b) Now consider Using the same concepts as in part (a) we

calculate and

16

and

Thus yielding:

From here we can calculate :

2.12 Consider the production function y = f(x). Since f exhibits constant returns to scale, f is homogenous of

degree 1. From Theorem 2.2, fi is homogenous of degree 0 (r=0), for any good i. Applying this and

Theorem 2.1 to fi ,

Furthermore, since f exhibits diminishing marginal productivity, fii < 0 for all i. In order for the above

condition to hold, there must exist a good j, such that fij >0. This implies that goods i and j are

complements in production.

2.13

2.14

2.15

2.16

2.17

2.18

2.19

17

2.20

2.21 We are given that individual A’s utility curve is defined by and

The proof of this will be complete when it is shown that

Therefore we calculate each of these and get

and

And using the fact that and we get:

which reduces to

2.22 We are given that , and we know the ATC is subaddative if .

Let , then . Also, , alternatively written

. But and .

Thus .

Recall that , then this can be re-written

.

2.23

CHAPTER THREE

3.1 and

So and

Therefore, -100<α<100.

3.2 To start with consider the supply and demand functions for the model, and

Therefore,

(a) Positive, because , and

18

(b) Negative.

(c) Negative.

(d) Positive.

3.3

3.4 Consider the two functions that characterize the short-run, perfectly competitive firm

(3.36)

(3.37)

The function (3.36) implies that , which can be substituted into (3.37) for the variable L.

Therefore (3.37) becomes and finding the profit maximizing condition yields

. This yields and which equals the conditions (3.42) and

(3.43) and confirms the answer.

3.5

3.6

3.7 (a)

(b) We need to find . We know .

19

3.8 Consider the function . The second order conditions imply that , , and

.

But and since labor and capital inputs cannot be negative. This

implies that the second order conditions do not hold.

3.9

3.10 Consider an imperfectly completive firm with total revenue, TR = TR(y, R) = 2yR + 116R, and total cost,

TC = TC(y, R) = y2 + 4y + 5R2. Therefore, the profit function π = π(y,R) = TR – TC = 2yR + 116R – y2 –

4y -5R2.

(a) To solve the first-order conditions for profit maximization, we first taking the partial derivative of π

with respect to y and R, yielding πy = 2R – 2y – 4 and πR = 2y + 116 – 10R. Setting πy and πR equal to

0, yields the following profit maximizing output y and market research expenditure R, y* = R – 2 and

R* = (y+58)/2

(b) The second order conditions are: πyy < 0, πRR < 0, and πyyπRR – (πyR)2 > 0. Using the answer to (a) we

have, πyy=–2, πRR=–10, and πyyπRR – (πyR)2 = (–2) (–10) – 2 = 18. So the second order conditions hold.

3.11

3.12

3.13

3.14

3.15

20

3.16

3.17

3.18

3.19

3.20

3.21

3.22

3.23

3.24 (a) Using the equations given:

with found in the exact same manner.

(b) Using the profit function for firm 1, ,

Solving for we obtain the following function:

and without loss of generality,

Now plugging in into yields:

and plugging that into yields the same thing.

21

(c)

3.25 Consider the Stackelberg leadership model where firm 1 is the leader. This model tells us that firm 1 will

maximize profits based on q2 = R2(q1) = (b – q1)/2. From (3.86) we have the following profit function for

firm 1:

The optimization problem becomes:

Max

The first order condition is:

The solution is and

3.26

3.27

3.28

CHAPTER FOUR

4.1 We know from (4.3) that

22

(a) Since MC(y) = 50 + 20y – 9y2, we have TC(y) = 50y + 10y2 – 3y3 + C. Using the initial value

condition TC(5) = 700, we have TC(y) = 575 + 50y + 10y2 – 3y3.

(b) Since MC(y) = 75 - 36y + 12y2, we have TC(y) = 75y - 18y2 + 4y3 + C. Using the initial value

condition TC(10) = 3000, we have TC(y) = 50 + 75y - 18y2 + 4y3.

4.2

4.3 The choice problem is subject to px + y = M, which implies y = M – px. Using this along with the original

maximization problem, we have, Max{x} U(x) = v(x) + (M – px). Thus, the first order condition is: v’(x) –

p(x) = 0. This implies: v’(x) = p(x). So we have the following:

And with a little algebra, this becomes U(x0) = M + CS(x0).

4.4

4.5

4.6

4.7

4.8

4.9 The problem supplies all the crucial information in order to compute the present value.

(a) , , , and

Using formula (4.42) the present value can be calculated which equals the maximum paid for the bond

(b) same as above but

23

4.10

4.11 Machine A has a 10 year life span and generates profits of $10,000/year

Machine B has a 6 year life span and generates profits of $10,000/year and a scrap value.

Assume .

In order to find the scrap value the difference between the machines’ present values needs to be calculated.

4.12

4.13

4.14

4.15

4.16

4.17

4.18 First, is the quantity of lumber available, and denote and .

We know that and we differentiate with respect to r to obtain

24

Since we know . Thus is negative. So, <0.

This means as interest rates rise, the optimal time to cut down the forest decreases.

4.19 Using discrete time analysis, we first form an equation for the present value of profit, :

For the firm to maximize its profit from this machine, it must maximize the amount of durability it is going

to obtain from the machine. Therefore, the cost of the marginal durability must equal that of the revenue

obtained per year of the machine:

4.20

4.21

4.22

4.23 (a) Plugging in into the given equation gives us .

Thus, . Solving for n* then gives the following equation:

(b) Using the equation from (a) yields 8 firms would be needed to maximize the net benefit to society.

CHAPTER FIVE

5.1

5.2

5.3

25

5.4

5.5

5.6

5.7

5.8

5.9

5.10

5.11

5.12

5.13

5.14

5.15

5.16

5.17

5.18

CHAPTER SIX

6.1

6.2

26

6.3 Consider the function with and constant returns to scale. This implies that the function

exhibits homogeneity of degree 1. By Theorem 2.2, it is known that exhibits homogeneity of

degree 0 and therefore and the determinant implying that the Hessian is singular.

6.4

6.5 (a) Plugging in values given into the IS-LM on page 97 yields:

Rearranging these statements yields:

Now, using Cramer’s Rule, we derive and .

First :

is obtained through a similar process:

(b) Doing direct differentiation on the answers in (a) we obtain the following:

27

(c) They are consistent.

6.6

6.7 Since gross substitution means , then all elements of . So there exist unique

differentiable inverse functions by Theorem 6.1

6.8

6.9

6.10

6.11 (a) Recall that

Therefore, using the labor coefficients given:

(b) Now, recall that and using equation 6.47:

(c) Repeating the process of part (b) with the new labor coefficients yield:

Thus, the price of the labor intensive good increased.

28

6.12 (a) From equation (6.30), x* = (I-A)-1d. So then we have,

(b) From equation (6.47), p* = waL(I-A)-1. So then we have,

6.13 We are given that A is productive, i.e. .

We will use Cramer’s Rule to find and show it is strictly greater than 0, and then find to show it is

greater than or equal to 0.

or written in matrix form,

Now we use Cramer’s Rule to find .

Here, we know the numerator is positive because the leading principle minor of n-1 degrees is positive.

The denominator is a P matrix by Theorem 6.3, and thus the determinant is positive. Therefore, is

strictly positive.

Now each because we do not know terms, thus if there are no 0 terms, the principle minors are

positive. If there are 0 terms, then .

29

6.14 We first note that Then using equations (6.43), V=p(I-A), and (6.30), x=(I-A)-1d:

So, This means that the total value added for each good is equal to the sum of the

price times the final consumption demand of each good. Or put another way, the total value added for each

good is equal to the final consumption revenue for each good.

6.15

6.16 Consider an additively separable utility function exhibiting diminishing marginal

utility for all goods . By Theorem 6.5 the Hessian needs to be symmetric and be an NP

matrix in order to be negative definite. In the case of this utility function the matrix looks like

with n rows and columns. The diagonal is negative due to the property of exhibiting

marginal utility and the off diagonal entries are all zero because of the utility function being additively

separable. This matrix is symmetric and also qualifies as a NP matrix because the determinant is simply

the product of the diagonal entries. By taking the principal minors the signs alternate and therefore the

Hessian is negative definite.

6.17

6.18

CHAPTER SEVEN

30

7.1 Consider the coefficient matrix and , which form the column

vector . Using (7.4) we determine and using

Cramer’s rule we can determine .

7.2

7.3

7.4

7.5

7.6

7.7 We first note that by equation (7.48). Now

taking the derivative of this with respect to p, we have,

And then by equation (7.37) we know , for all i. So we have,

.

7.8

31

7.9 By (7.53) it is known that

Therefore

Also by (7.53) it is known that

Therefore

The order of differentiation does not matter, and therefore

validation the condition.

7.10

7.11

7.12

7.13

7.14

7.15 (a)

Now differentiate with respect to a and b to obtain first-order conditions:

32

(b)

(c)

7.16 Recall that:

Using the given equations for profit for each of the respective companies, start by finding and :

33

Now, solve for and for :

And then plug into and resolve for :

Here we can impose symmetry for . Now, using the tax revenue function :

To find and , find since we can impose symmetry:

Solve for :

7.17

7.18

CHAPTER EIGHT

8.1

8.2

34

8.3

8.4 Consider the function

Subject to

(a) The Lagangrian is

And the corresponding first-order conditions are

Solving for and yields and

(b) The answer is the same as (8.2) just with added a parameters that unequally distribute the products

whereas in (8.2) only price determines the distribution of the two goods.

8.5

8.6

8.7

8.8

8.9 Consider an economy that produces only two goods, x and y. These goods are produced using labor L

input according to the production functions and .

(a) Differentiating these equations with respect to their labor gives us the following marginal return

functions: and .

Differentiating these again with respect to there labor gives, and

35

. Since labor L must be nonnegative, these are always negative

and thus exhibit diminishing marginal returns.

(b) Solving the production functions for labor, we have and . Putting these into the

constraining labor condition (Lx + Ly = 1), we have . Which simplifies to

.

(c) This is simply the equation for a circle of radius 2 centered at the origin. When we restrict our

ourselves to x,y>0 we have a quarter-circle that has the standard concave shape.

(d) We want to maximize xy, subject to . Finding the Lagrangian, we have

. This yields the following first order condition:

Solving this system for x* and y* yields, .

8.10

8.11 (a)

(b)

(c) Differentiate each first order condition with respect to r:

36

(d)

In interest rises, you consume less in the present day and consume more later.

8.12

8.13

8.14

8.15 (a) Using the specific utility function set up in the text; the dual problem is set up as follows:

Min

Subject to:

From this, the Lagrangian is:

The first order conditions are then:

(1)

(2)

(3)

Solving (1) for yields , which then can be used in equation (2) to get .

Using in equation (3) gives us an equation for :

37

Taking this result and plugging it back into our equation for yields the following:

(b) Yes they are symmetric, as it has already been shown.

(c) To find the expenditure function, we utilize the following formula:

Simplifying this yields:

(d) Recall from the text that and . Using for M gives us

the following two equations:

Thus, for .

(e) Recall, to calculate the cross effects for this problem, we need to utilize the following expression:

Now find the specific derivatives required:

To solve, plug these back into the original expression to obtain:

38

Now to calculate the own effects, we utilize the following equation:

After finding the appropriate derivatives and plugging them into the above expression, we obtain the

following:

This holds, thus implying the Slutsky equations for this demand function holds for good 1.

8.16 (a) Consider the function

Subject to the

The Langragian is

and the corresponding fist-order conditions are

Solving for and yields

(b) Consider the function

Subject to the

The Langragian is

and the corresponding fist-order conditions are

Solving for and yields and

39

(c)

As long as individuals will be better of because or in other words as

long as the population is growing, individuals will be better off with the social security system.

(d) Yes, the government will be able to finance this “pay as you go” social security system, because

and

We also know that and therefore Borrowed=Paidout

8.17

8.18 (a)

To satisfy all second order conditions, all leading principle minors must be negative:

(b) Prove MC increasing when w increasing, i.e.

40

since

8.19 Consider a consumer with fixed money income M, who purchases goods x1 and x2 so as to maximize the

utility function , where a and b are positive constants.

(a) Finding the Lagrangian, we have . This

yields the following first order condition:

Solving this system yields: , , and .

(b) Finding the bordered Hessian,

Then we have the following principle minors,

41

So the utility function satisfies the second order conditions.

(c) The marginal utility of money income is given by . So differentiating with respect to pi and M

yields,

So the marginal utility of money income is constant with respect to prices but not money income.

8.20

8.21

8.22

8.23

8.24

8.25

8.26 In this problem we want to show that . For this, we are simply going to prove for the

case of j=1,2 since the steps would be similar to proving the identity in general. Now, recall that

. From this expression we know the following:

(1)

(2)

To prove that (2) is true, consider the following claim:

42

Since we are only considering this for n=2 cases, the proof looks as follows:

Therefore, returning to expression (2), we see this result proves the equality. Using the results of (1) and

(2) we see that Roy’s Identity holds.

8.27

8.28

8.29

8.30

8.31

8.32

CHAPTER NINE

9.1

9.2

9.3 Consider the two-good consumer choice problem in (9.24). Assuming that U1 = U2 where U1,U2>0 and p2

> p1. There are the following four cases:

Case: x1*>0 and x2*>0. From (9.26b) we have and from (9.26d) we have . But this

implies . Since U1 = U2, but this implies p1 = p2. A contradiction.

43

Case: x1*=0 and x2*=0. From (9.26f), . But since M>0, this implies . Substituting

into (9.26a) implies . A contradiction.

Case: x1*=0 and x2*>0. From (9.26d), we have and from (9.26a) . These imply

and thus . But since U1 = U2, we have but this implies . A

contradiction.

Case: x1*>0 and x2*=0. From (9.26b), we have and from (9.26c) . These imply

and thus . Since U1 = U2, we have and thus , which is true.

Therefore it follows that x1*>0 and x2*=0.

9.4

9.5

From (9.37b), because .

From (9.37a),

From (9.37d),

44

Since , , and since , .

So,

Then .

We know in this production function, , thus without the constraint,

. Hence .

9.6

9.7

9.8

9.9

9.10

9.11 Consider the problem faced by a macroeconomic policy maker who can control the level of real national

income Y and the rate of inflation via various policy instruments.

(a) The short-run Phillips curve shows the trade off between unemployment and inflation. If one were to

graph it in space, it would be downward slopping curve that bows inward towards the origin.

(b) Given our maximization problem, we have the following Lagrangian and Kuhn-Tucker Conditions:

(c) Given , we have the following Lagrangian and Kuhn-Tucker Conditions:

45

Given , we have and thus . And given , we have

and thus . And then since we have

and thus .

(d) Differentiating our results from (c) and given that and , we have the following signed

comparative statics terms: , ,

, , , .

(e) All of the comparative statics terms in (d) make economic sense. Their interpretations are as follows:

- as increases, Y* decreases or as the excepted rate of inflation increases, output

decreases, - as increases, Y* increases or as the responsiveness to inflation increases,

output increases, - as increases, Y* increases equal proportionally or as the full-

employment level increases, output increases equal proportionally, - there is no

relationship between and or there is no relationship between the expected rate of inflation

and the actual rate of inflation, - as increases, increases or as the responsiveness

to inflation increases, the actual rate of inflation increases, - there is no relationship

between and or there is no relationship between the full-employment level and the actual

rate of inflation.

9.12

46

9.13

9.14

9.15 Consider the GDP maximization problem from (9.47).

(9.56) Gives the formula to set up the corresponding dual.

Subject to:

From (9.47) we know that

,Total GDP=76, and

In this case the subscript i’s in (9.56) correspond to the 3 input T,L,K and by knowing total GDP the

inequalities can be changed to equalities. Therefore (9.56) becomes

Subject to:

Solving for yields , , and .

9.16

9.17 In this problem we want to maximize the following utility function with the usual budget constraint:

Subject to:

The Lagrangian is then set up as follows with its first order conditions:

(1)

47

(2)

(3)

(4)

(5)

(6)

(a) and

Under these conditions, using equation (2) and , we only have to solve for using the fact

that :

From this we get our first restriction: . Using this we can then say that , which then

allows us to examine equation (5) as an equality:

(b) and

Utilizing the same process in (a) here, but instead using equation (4), we see that equation (3) becomes

an equality statement, which can be solved for :

From this we then get our next restriction: . This then allows us to say that , which

turns equation (5) into an equality that can be solved for :

(c) and

Under these new conditions, equations (1), (3) and (5) for the first order conditions of the Lagrangian

can be made equalities. Solving equation (1) for yields:

48

Using this, solve equation (3) for :

Now take this result and solve equation (5) for :

Here we can impose symmetry and say that .

9.18

49