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To solve this problem, let’s go through each question carefully.

### Problem Analysis
The steel manufacturer has:
- **Production function** $ P(K, L) $ in terms of capital $ K $ and labor $ L $.
- **Production cost function** $ C(K, L) $ with a budget constraint of $800,000.
- Information given:
  - Maximum production = 10,000 tons
  - Capital cost = $550,000
  - Labor cost = $250,000
- **Lagrange multiplier** $ \lambda = 0.12 $.

Let’s use this information to answer each question.

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### Part (a): Objective Function
The objective function generally represents what the manufacturer wants to **maximize** or **minimize**. Here, it’s likely the **production output** $ P(K, L) $, as the manufacturer aims to maximize steel production within the budget constraint.

**Answer for (a):** The objective function is $ P(K, L) $.

---

### Part (b): Constraint
The constraint represents the **budget limit** on production costs, which should not exceed $800,000.

The cost function $ C(K, L) $ is given by the capital and labor costs, so the constraint is:
\[
C(K, L) = 800,000
\]

**Answer for (b):** $ C(K, L) = 800,000 $

---

### Part (c): Units for $ \lambda $
The Lagrange multiplier $ \lambda $ indicates the **rate of change of the objective function** (production) with respect to the constraint (budget). In this case:
- Since production is measured in **tons** and the constraint is in **dollars**, $ \lambda $ has units of **tons per dollar**.

**Answer for (c):** Tons per dollar.

---

### Part (d): Practical Meaning of $ \lambda = 0.12 $
The value $ \lambda = 0.12 $ suggests that **for each additional dollar** added to the budget, the production increases by approximately **0.12 tons**.

**Answer for (d):** Every extra dollar of budget increases maximal production by approximately $ 0.12 $ tons.

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### Summary of Answers:
1. **(a)** Objective function: $ P(K, L) $
2. **(b)** Constraint: $ C(K, L) = 800,000 $
3. **(c)** Units of $ \lambda $: Tons per dollar
4. **(d)** Practical meaning: Every extra dollar of budget increases maximal production by approximately 0.12 tons.

Would you like further explanation on any part of this solution, or additional details about Lagrange multipliers?

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### Related Questions
1. What does the Lagrange multiplier indicate in optimization problems?
2. How does one derive the objective function and constraints in production economics?
3. Can the Lagrange multiplier’s value vary if other constraints are added?
4. How would the solution change if the budget constraint was set to $900,000 instead?
5. In practical terms, how can a company decide how much labor and capital to allocate given budget constraints?

### Tip
When dealing with optimization problems with constraints, setting up the objective function and understanding constraints clearly helps in applying techniques like the Lagrange multiplier method efficiently.

A steel manufacturer can produce P(K, L) tons of steel using K units of capital and L units of labor, with production costs C(K, L) dollars. With a budget of $800,000, the maximum production is 10,000 tons, using $550,000 of capital and $250,000 of labor. The Lagrange multiplier is λ = 0.12.
(a) What is the objective function?
(b) What is the constraint?
(c) What are the units for λ?
(d) What is the practical meaning of the statement λ = 0.12?

Learn how to solve an optimization problem involving steel production with budget constraints using the Lagrange multiplier method. This example includes calculating objective functions, constraints, and understanding the practical meaning of λ in production increase.

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