Jesaki Furniture Company has two plants that produce lumber used in manufacturing tables and chairs. In 1 day of operation, plant A can produce the lumber required to manufacture 21 tables and 56 chairs, while plant B can produce the lumber required to manufacture 28 tables and 49 chairs. In this production run, the company needs enough lumber to manufacture at least 216 tables and 509 chairs. If it costs $1,086 per day to operate plant A and $917 per day to operate plant B, what is the minimum cost of this production run? $ Round to the nearest cent.

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### Jesaki Furniture Company Production Cost Optimization

**Problem Statement:**

Jesaki Furniture Company has two plants that produce lumber used in manufacturing tables and chairs. In 1 day of operation, plant A can produce the lumber required to manufacture 21 tables and 56 chairs, while plant B can produce the lumber required to manufacture 28 tables and 49 chairs. In this production run, the company needs enough lumber to manufacture at least 216 tables and 509 chairs.

If it costs $1,086 per day to operate plant A and $917 per day to operate plant B, what is the minimum cost of this production run?

**Mathematical Formulation:**

1. Let \( x \) be the number of days plant A operates.
2. Let \( y \) be the number of days plant B operates.

Given:
- Plant A: \( 21x \) tables and \( 56x \) chairs
- Plant B: \( 28y \) tables and \( 49y \) chairs

Constraints:
1. \( 21x + 28y \geq 216 \) (tables requirement)
2. \( 56x + 49y \geq 509 \) (chairs requirement)

Cost Function:
\[ \text{Total Cost} = 1086x + 917y \]

**Objective:**

Minimize the total cost subject to the given constraints.

**Solution:**

Using linear programming tools or graphical methods, you can solve the above linear inequalities to find the values of \( x \) and \( y \) that minimize the total cost.

\[ \$ ______________. \] 

**Note:** Round the final cost to the nearest cent.

**Example Calculation:**

Let's assume through solving the inequalities, we find \( x = 6 \) (plant A operates for 6 days) and \( y = 4 \) (plant B operates for 4 days).

Then, the total cost = \( 6 \times 1086 + 4 \times 917 = 6516 + 3668 = \$ 10184 \).

Hence, the minimum cost could be annotated as:
\[ \$ 10184.00 \] (rounded to the nearest cent)

The exact values can be derived from precise linear programming solutions.
Transcribed Image Text:### Jesaki Furniture Company Production Cost Optimization **Problem Statement:** Jesaki Furniture Company has two plants that produce lumber used in manufacturing tables and chairs. In 1 day of operation, plant A can produce the lumber required to manufacture 21 tables and 56 chairs, while plant B can produce the lumber required to manufacture 28 tables and 49 chairs. In this production run, the company needs enough lumber to manufacture at least 216 tables and 509 chairs. If it costs $1,086 per day to operate plant A and $917 per day to operate plant B, what is the minimum cost of this production run? **Mathematical Formulation:** 1. Let \( x \) be the number of days plant A operates. 2. Let \( y \) be the number of days plant B operates. Given: - Plant A: \( 21x \) tables and \( 56x \) chairs - Plant B: \( 28y \) tables and \( 49y \) chairs Constraints: 1. \( 21x + 28y \geq 216 \) (tables requirement) 2. \( 56x + 49y \geq 509 \) (chairs requirement) Cost Function: \[ \text{Total Cost} = 1086x + 917y \] **Objective:** Minimize the total cost subject to the given constraints. **Solution:** Using linear programming tools or graphical methods, you can solve the above linear inequalities to find the values of \( x \) and \( y \) that minimize the total cost. \[ \$ ______________. \] **Note:** Round the final cost to the nearest cent. **Example Calculation:** Let's assume through solving the inequalities, we find \( x = 6 \) (plant A operates for 6 days) and \( y = 4 \) (plant B operates for 4 days). Then, the total cost = \( 6 \times 1086 + 4 \times 917 = 6516 + 3668 = \$ 10184 \). Hence, the minimum cost could be annotated as: \[ \$ 10184.00 \] (rounded to the nearest cent) The exact values can be derived from precise linear programming solutions.
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