Chapter 6 Henderson and Quandt Microeconomics, Questions 6.9-6.11_ Solutions

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6-9 A consumer allocates a fixed amount of time to labor and leisure. He derives satisfaction

from the time he retains as leisure, L, and the income, y, that he secures by selling his labor at a

fixed wage rate. His utility function is U = Ly+aL where a is a positive parameter. Derive the

consumer's supply function for labor. Is his labor supply curve upward-sloping?

The consumer’s utility function is U = Ly + aL, where L is leisure time and y is income earned from selling labor at a fixed wage rate.

The consumer’s budget constraint is:

Y = w(24 – L)

Where w is the fixed wage rate and 24 is the total amount of time available in a day.

The consumer’s problem is to maximize utility subject to the budget constraint:

Max U = Ly + aL

s.t. y = w(24 – L)

The Lagrangian for this problem is:

L = Ly + aL + λ(y – w(24 – L))

Taking partial derivatives with respect to L, y, and λ, we get:

dL/dL = y + a – λw = 0

dL/dy = L – λ = 0

dL/dλ = y – w(24 – L) = 0

From the second equation, we get:

Λ = L

Substituting this into the first equation, we get:

Y + a = wL

Solving for L, we get:

L = (y + a)/w

This is the consumer’s labor supply function. Since L is positively related to y, the labor supply curve is upward sloping. This means that as the wage rate increases, the consumer will supply more labor, and as the wage rate decreases, the consumer will supply less labor.

6-10 Assume that aggregate demand and supply functions are given by D=25/p and S = √5p. Is

the dynamic process defined by (6-21) locally stable?

The dynamic process referred to in (6-21) is a price adjustment process where the current price (p_t) is adjusted towards the expected long-run equilibrium price (p*) according to the following equation:

P_t+1 = (1 – α) p_t + α p*

Where α is a constant parameter between 0 and 1.

To determine the stability of this dynamic process, we need to calculate the derivative of p_t+1 with respect to p_t and evaluate it at the steady state solution (p*). If the absolute value of this derivative is less than 1, then the dynamic process is locally stable. Otherwise, it is unstable.

First, we need to find the long-run equilibrium price by setting aggregate demand equal to aggregate supply:

25/p* = √5p*

Solving for p*, we get:

P* = 5

Next, we need to calculate the derivative of p_t+1 with respect to p_t:

D(p_t+1)/dp_t = 1 – α + α d(p*/p_t)/dp_t

Substituting the values of p* and the aggregate supply function, we get:

D(p_t+1)/dp_t = 1 – α + α (-√5/p_t)

Evaluating this expression at p* = 5, we get:

D(p_t+1)/dp_t|p*=5 = 1 – α – α/√5

For the dynamic process to be locally stable, we need:

|d(p_t+1)/dp_t| < 1

|1 – α – α/√5| < 1

Solving this inequality, we get:

0 < α < 2/(2 + √5)

Since α is between 0 and 1 by definition, the inequality is satisfied, and the dynamic process defined by (6-21) is locally stable.

6-11 Determine whether equilibrium solutions exist for markets with the following demand and

supply

(a) D=12-3p; S = -10+2p.

(b) D=16-2p; S = 20-2p.

(c) D=50-4p; S = 10 + 10p-p².

(d) D=50-4p; S = 2+10p-p².

(a) To find equilibrium solutions, we need to set the demand equal to the supply:

12-3p = -10+2p

Simplifying the above equation, we get:

5p = 22

P = 4.4

Therefore, equilibrium solution exists for this market.

(b) To find equilibrium solutions, we need to set the demand equal to the supply:

16-2p = 20-2p

Simplifying the above equation, we get:

0 = 0

This means that for any value of p, the demand is equal to the supply. Therefore, this market has an infinite number of equilibrium solutions.

(c) To find equilibrium solutions, we need to set the demand equal to the supply:

50-4p = 10 + 10p-p²

Simplifying the above equation, we get:

P³ - 14p² + 60p – 40 = 0

Factorizing the above equation, we get:

(p-2)(p-4)(p-5) = 0

Therefore, the equilibrium solutions are p=2, p=4, and p=5.

(d) To find equilibrium solutions, we need to set the demand equal to the supply:

50-4p = 2+10p-p²

Simplifying the above equation, we get:

P³ - 14p² + 48p – 48 = 0

Factorizing the above equation, we get:

(p-2)(p-4)(p-6) = 0

Therefore, the equilibrium solutions are p=2, p=4, and p=6.

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