To find the maximum value of A at which the consumer buys 10 units or less, we need to determine the utility maximizing amount of good x that the consumer will purchase.
First, let's break down the pricing rule. The consumer faces a piecewise linear price function:
- For the first 10 units, the price per unit is $2.
- For any additional units beyond 10, the price per unit is $3.
We can denote the quantity of good x purchased as q. The consumer's budget constraint can be written as follows:
2q ≤ m
where m represents the consumer's budget.
To determine the consumer's optimal choice, we need to express their utility function in terms of q. Given the utility function U(x, m) = A ln(2x) + m, we substitute q for x:
U(q, m) = A ln(2q) + m
Now, we need to determine the maximum amount of x that the consumer will buy. This will depend on the consumer's budget constraint and their utility maximization problem.
Using Lagrange multipliers, we set up the following problem:
Maximize U(q, m) subject to the constraint 2q ≤ m.
To solve this problem, we differentiate the utility function with respect to q and set it equal to the Lagrange multiplier (λ) times the derivative of the budget constraint with respect to q. The equation becomes:
∂U/∂q = λ * ∂(2q)/∂q
Differentiating the utility function gives us:
A * 2/2q = λ * 2
Simplifying:
A/q = λ
From the budget constraint, we know that 2q = m. Substituting this into the equation, we get:
A/(m/2) = λ
Simplifying further:
2A/m = λ
Rearranging the equation, we can express A in terms of λ and m:
A = (λ * m)/2
Now, we need to solve for the values of λ and m that satisfy the budget constraint. Since the consumer will buy a maximum of 10 units, we have the following condition:
10 ≤ q ≤ m/2
Substituting q = 10, we get:
10 ≤ 10A
This simplifies to:
1 ≤ A
Therefore, the maximum value of A at which the consumer buys 10 units or less is A = 1. If A exceeds 1, the consumer's optimal choice will exceed 10 units.
I hope this explains how to approach this problem. Let me know if you need any further clarification!