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Sagot :
To solve this problem, we need to maximize Andrew's profit given the constraints on resources and production. Let's start by defining the problem mathematically, then solve it step by step.
1. Variables and Objective Function:
- Let [tex]\( x \)[/tex] represent the number of bookcases.
- Let [tex]\( y \)[/tex] represent the number of TV stands.
- The profit function [tex]\( P \)[/tex] that we need to maximize is:
[tex]\[ P = 60x + 100y \][/tex]
2. Constraints:
- Total number of furniture pieces: Andrew can make a maximum of 18 pieces of furniture.
[tex]\[ x + y \leq 18 \][/tex]
- Non-negative production: Andrew cannot produce a negative number of furniture pieces.
[tex]\[ x \geq 0 \][/tex]
[tex]\[ y \geq 0 \][/tex]
- Material cost constraint: The cost to make bookcases and TV stands should not exceed \[tex]$600.00. \[ 50x + 100y \leq 600 \] 3. Formulating the inequalities: - Total number of furniture pieces: \[ x + y \leq 18 \] - Cost constraint: \[ 50x + 100y \leq 600 \] - Non-negativity: \[ x \geq 0 \] \[ y \geq 0 \] 4. Graphical Solution: To find the feasible region, plot the following on a graph: - \( x + y = 18 \) - \( 50x + 100y = 600 \) - \( x = 0 \) - \( y = 0 \) Simplifying the inequality \( 50x + 100y \leq 600 \): \[ x + 2y \leq 12 \] 5. Vertices of the Feasible Region: To find the vertices, solve the system of inequalities: - Point A: Intersection of \( x + y = 18 \) and \( x + 2y = 12 \) \[ x + y = 18 \quad \text{(1)} \] \[ x + 2y = 12 \quad \text{(2)} \] Subtract equation (2) from (1): \[ x + y - (x + 2y) = 18 - 12 \] \[ -y = 6 \implies y = -6 \quad \text{(Not feasible within the constraints)} \] - Point B: Intersection of \( x = 0 \) and \( x + 2y = 12 \) \[ 0 + 2y = 12 \implies y = 6 \] So, one vertex is \( (0, 6) \). - Point C: Intersection of \( y = 0 \) and \( x + 2y = 12 \) \[ x + 2(0) = 12 \implies x = 12 \] So, another vertex is \( (12, 0) \). - Point D: Intersection of \( x = 0 \) and \( x + y = 18 \) \[ 0 + y = 18 \implies y = 18 \] But this point does not satisfy the cost constraint. 6. Evaluating Profit at Vertices: - For point \( (0, 6) \): \[ P = 60(0) + 100(6) = 600 \] - For point \( (12, 0) \): \[ P = 60(12) + 100(0) = 720 \] 7. Determining Maximum Profit: The maximum profit is obtained at the vertex \( (12, 0) \), where the profit \( P \) is \$[/tex]720.00.
Conclusion:
Andrew should make 12 bookcases and 0 TV stands to maximize his profit. The maximum achievable profit in this situation is \$720.00.
1. Variables and Objective Function:
- Let [tex]\( x \)[/tex] represent the number of bookcases.
- Let [tex]\( y \)[/tex] represent the number of TV stands.
- The profit function [tex]\( P \)[/tex] that we need to maximize is:
[tex]\[ P = 60x + 100y \][/tex]
2. Constraints:
- Total number of furniture pieces: Andrew can make a maximum of 18 pieces of furniture.
[tex]\[ x + y \leq 18 \][/tex]
- Non-negative production: Andrew cannot produce a negative number of furniture pieces.
[tex]\[ x \geq 0 \][/tex]
[tex]\[ y \geq 0 \][/tex]
- Material cost constraint: The cost to make bookcases and TV stands should not exceed \[tex]$600.00. \[ 50x + 100y \leq 600 \] 3. Formulating the inequalities: - Total number of furniture pieces: \[ x + y \leq 18 \] - Cost constraint: \[ 50x + 100y \leq 600 \] - Non-negativity: \[ x \geq 0 \] \[ y \geq 0 \] 4. Graphical Solution: To find the feasible region, plot the following on a graph: - \( x + y = 18 \) - \( 50x + 100y = 600 \) - \( x = 0 \) - \( y = 0 \) Simplifying the inequality \( 50x + 100y \leq 600 \): \[ x + 2y \leq 12 \] 5. Vertices of the Feasible Region: To find the vertices, solve the system of inequalities: - Point A: Intersection of \( x + y = 18 \) and \( x + 2y = 12 \) \[ x + y = 18 \quad \text{(1)} \] \[ x + 2y = 12 \quad \text{(2)} \] Subtract equation (2) from (1): \[ x + y - (x + 2y) = 18 - 12 \] \[ -y = 6 \implies y = -6 \quad \text{(Not feasible within the constraints)} \] - Point B: Intersection of \( x = 0 \) and \( x + 2y = 12 \) \[ 0 + 2y = 12 \implies y = 6 \] So, one vertex is \( (0, 6) \). - Point C: Intersection of \( y = 0 \) and \( x + 2y = 12 \) \[ x + 2(0) = 12 \implies x = 12 \] So, another vertex is \( (12, 0) \). - Point D: Intersection of \( x = 0 \) and \( x + y = 18 \) \[ 0 + y = 18 \implies y = 18 \] But this point does not satisfy the cost constraint. 6. Evaluating Profit at Vertices: - For point \( (0, 6) \): \[ P = 60(0) + 100(6) = 600 \] - For point \( (12, 0) \): \[ P = 60(12) + 100(0) = 720 \] 7. Determining Maximum Profit: The maximum profit is obtained at the vertex \( (12, 0) \), where the profit \( P \) is \$[/tex]720.00.
Conclusion:
Andrew should make 12 bookcases and 0 TV stands to maximize his profit. The maximum achievable profit in this situation is \$720.00.
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