(a) A firm operates with the production function Q = 4K^0.6 L^0.4 and buys inputs K and L at prices per unit of $40 and $15 respectively. What is the cheapest way of producing 600 units of out?
(b) A firm has a budget of $300 to spend on the three inputs x, y and z whose prices per unit are $4, $1, and $6 respectively. What combination of x, y, and z should it employ to maximize output if it faces the production function Q = 24x^0.3 y^0.2 z^0.3?
(b) A firm uses 200, 000 units of a component in a year, with demand evenly spread over the year. In addition to the purchase price, each order placed for a batch of compounds costs $80. Each unit held in stock over a year costs $8. What is the optimum order size?
(a) To find the cheapest way of producing 600 units of output, we need to minimize the cost of production. The cost function is given by the equation:
C = 40K + 15L
To produce 600 units of output, we have:
Q = 4K^0.6 * L^0.4 = 600
We can rewrite the production function as:
K = (600 / (4L^0.4))^1/0.6
Substituting the value of K into the cost function, we have:
C = 40(600 / (4L^0.4))^1/0.6 + 15L
To minimize the cost, we need to find the value of L that minimizes C. We can do this by taking the derivative of C with respect to L and setting it equal to zero:
dC/dL = -40(600 / (4L^0.4))^1/0.6 * (0.4L^-0.6) + 15 = 0
Simplifying this equation, we get:
-40(600 / (4L^0.4))^1/0.6 * 0.4L^-0.6 = -15
Solving for L, we find:
L = 155.083
Plugging this value of L back into the production function, we can find the value of K:
K = (600 / (4 * 155.083^0.4))^1/0.6 ≈ 11.146
Therefore, the cheapest way of producing 600 units of output is by using approximately 11.146 units of capital (K) and 155.083 units of labor (L).
(b) To find the combination of x, y, and z that maximizes output, subject to the budget constraint, we can use the method of Lagrange multipliers.
The constrained optimization problem can be set up as follows:
Maximize Q = 24x^0.3 * y^0.2 * z^0.3
Subject to the constraint: 4x + y + 6z = 300
The Lagrangian function is:
L = Q - λ(4x + y + 6z - 300)
To find the optimum combination, we need to take the partial derivatives of the Lagrangian function with respect to x, y, z, and λ, and set them equal to zero:
∂L/∂x = 0.3(24x^0.3 * y^0.2 * z^0.3) - 4λ = 0
∂L/∂y = 0.2(24x^0.3 * y^0.2 * z^0.3) - λ = 0
∂L/∂z = 0.3(24x^0.3 * y^0.2 * z^0.3) - 6λ = 0
∂L/∂λ = 4x + y + 6z - 300 = 0
Solving these equations simultaneously, we can find the values of x, y, z, and λ that satisfy the equations. These values will give us the optimum combination.
(c) To find the optimum order size, we need to minimize the total cost of ordering and holding stock. The cost function can be expressed as:
C = Q * (Purchase cost per unit + Cost per order) + Q * (Holding cost per unit)
In this case, the purchase cost per unit is given as $200, the cost per order is $80, and the holding cost per unit is $8.
Let's denote the order size as Q. The total cost function can be rewritten as:
C = 200Q + 80(200,000 / Q) + 8(200,000 / 2)
To find the optimum order size, we need to find the value of Q that minimizes C. We can do this by taking the derivative of C with respect to Q and setting it equal to zero:
dC/dQ = 200 - 80(200,000 / Q^2) = 0
Simplifying this equation, we get:
200Q^2 = 80(200,000)
Solving for Q, we find:
Q^2 = 80(200,000) / 200
Q^2 = 80,000
The optimum order size is the square root of 80,000:
Q = √80,000
Q ≈ 282.84
Therefore, the optimum order size is approximately 282.84 units.