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Let S represent the amount of steel produced (in tons). Steel production is related to the amount of labor used (L) and the amount of capital used (C) by the following function:

S=20L0.30C0.70
In this formula L represents the units of labor input and C the units of capital input. Each unit of labor costs $50, and each unit of capital costs$100.
Formulate an optimization problem that will determine how much labor and capital are needed to produce 50,000 tons of steel at minimum

User Xmarcos
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Final answer:

To determine the best production method, calculate the total cost for each method and choose the one with the lowest cost. Method 1 has the lowest cost and should be used to produce 50,000 tons of steel. Even if the cost of labor rises, Method 1 remains the most cost-effective choice.

Step-by-step explanation:

To determine the best production method, we need to calculate the total cost for each method and choose the one with the lowest cost. Let's start with the given information:

  • Method 1: 50 units of labor, 10 units of capital
  • Method 2: 20 units of labor, 40 units of capital
  • Method 3: 10 units of labor, 70 units of capital

Given that each unit of labor costs $100 and each unit of capital costs $400, we can calculate the cost for each method:

  1. Method 1: Cost = (50 * $100) + (10 * $400) = $5,000 + $4,000 = $9,000
  2. Method 2: Cost = (20 * $100) + (40 * $400) = $2,000 + $16,000 = $18,000
  3. Method 3: Cost = (10 * $100) + (70 * $400) = $1,000 + $28,000 = $29,000

Based on these calculations, Method 1 has the lowest cost and should be used to produce 50,000 tons of steel. If the cost of labor rises to $200 per unit, the updated costs will be:

  1. Method 1: Cost = (50 * $200) + (10 * $400) = $10,000 + $4,000 = $14,000
  2. Method 2: Cost = (20 * $200) + (40 * $400) = $4,000 + $16,000 = $20,000
  3. Method 3: Cost = (10 * $200) + (70 * $400) = $2,000 + $28,000 = $30,000

Again, Method 1 has the lowest cost even with the increased labor cost, and it should still be used to produce 50,000 tons of steel.

User Sandeep Vokkareni
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