Final answer:
To determine the best production method, we need to calculate the cost per unit of output for each method. The production function is q = 10 L^0.5 K^0.5. By comparing the cost per unit of output for each method, it is determined that Method 1 is the best production method, regardless of the cost of labor.
Step-by-step explanation:
To determine the best production method, we need to calculate the cost per unit of output for each method. Let's start by calculating the cost per unit of output for each method:
Method 1: Total cost = (50 units of labor x $100/unit) + (10 units of capital x $400/unit) = $5,000 + $4,000 = $9,000. Cost per unit of output = $9,000 / q = $9,000 / (10 L^0.5 K^0.5).
Method 2: Total cost = (20 units of labor x $100/unit) + (40 units of capital x $400/unit) = $2,000 + $16,000 = $18,000. Cost per unit of output = $18,000 / q = $18,000 / (10 L^0.5 K^0.5).
Method 3: Total cost = (10 units of labor x $100/unit) + (70 units of capital x $400/unit) = $1,000 + $28,000 = $29,000. Cost per unit of output = $29,000 / q = $29,000 / (10 L^0.5 K^0.5).
The best production method is the one with the lowest cost per unit of output. To compare the costs for different levels of labor and capital, we need to evaluate the cost per unit of output as a function of these variables. But since the production function is given as q = 10 L^0.5 K^0.5, we can substitute this expression into the cost per unit of output equations and determine the method with the lowest cost:
Method 1: Cost per unit of output = $9,000 / (10 L^0.5 K^0.5) = $900 / (L^0.5 K^0.5) = $900 / (LK)0.25.
Method 2: Cost per unit of output = $18,000 / (10 L^0.5 K^0.5) = $1,800 / (L^0.5 K^0.5) = $1,800 / (LK)0.25.
Method 3: Cost per unit of output = $29,000 / (10 L^0.5 K^0.5) = $2,900 / (L^0.5 K^0.5) = $2,900 / (LK)0.25.
Comparing the cost per unit of output for each method, we can see that Method 1 has the lowest cost per unit of output regardless of the cost of labor. Therefore, the company should use Method 1 as the best production method.