IB MYP 4-5 Maths- Linear programming, including inequalities- Study Notes - New Syllabus
IB MYP 4-5 Maths- Linear programming, including inequalities – Study Notes
Extended
- Linear programming, including inequalities
IB MYP 4-5 Maths- Linear programming, including inequalities – Study Notes – All topics
Linear Programming and Inequalities
Linear Programming
Linear Programming (LP) is a method used to find the maximum or minimum value of an objective function (usually profit or cost) subject to a set of constraints expressed as linear inequalities.
Key Elements:
- Decision Variables: The unknowns we want to determine (e.g., number of products).
- Objective Function: The function to optimize (maximize or minimize), usually of the form: \( Z = ax + by \), where \( x, y \) are decision variables.
- Constraints: A set of linear inequalities representing limitations (e.g., time, resources).
- Feasible Region: The region on a graph where all constraints overlap.
- Optimal Solution: The point in the feasible region that gives the best value of the objective function (usually at a corner point).
Steps to Solve Linear Programming Problems:
- Define decision variables.
- Write the objective function.
- Write constraints as linear inequalities.
- Graph the inequalities to find the feasible region.
- Find the corner (vertex) points of the feasible region.
- Evaluate the objective function at each corner point.
- Select the maximum or minimum value as required.
Important Notes on Inequalities in Graphs:
- \( y \le mx + c \): Shade below the line.
- \( y \ge mx + c \): Shade above the line.
- Dashed line for strict inequalities (\( < \) or \( > \)).
- Solid line for inclusive inequalities (\( \le \) or \( \ge \)).
Example:
A company makes two products: A and B. Each unit of A requires 2 hours of labor and 3 kg of material. Each unit of B requires 1 hour of labor and 2 kg of material. The company has at most 8 hours of labor and 10 kg of material. Profit is $5 per unit of A and $4 per unit of B. Find how many units of each product should be made to maximize profit.
▶️ Answer/Explanation
Step 1: Define Variables
Let \( x \) = units of A, \( y \) = units of B.
Step 2: Objective Function
Maximize \( P = 5x + 4y \).
Step 3: Constraints
- Labor: \( 2x + y \le 8 \).
- Material: \( 3x + 2y \le 10 \).
- Non-negativity: \( x \ge 0, y \ge 0 \).
Step 4: Find Feasible Region & Corner Points
Vertices: (0,0), (0,5), (4,0), (2,4).
Step 5: Evaluate Profit
- (0,0): \( P=0 \)
- (0,5): \( P=20 \)
- (4,0): \( P=20 \)
- (2,4): \( P=26 \)
Optimal Solution: \( x=2, y=4 \), Maximum profit = $\$26.$
Example:
A farmer can plant at most 100 acres of crops: wheat and corn. Each acre of wheat requires 3 hours of labor, and each acre of corn requires 2 hours of labor. The farmer has at most 240 hours of labor. Profit is $\$200$ per acre of wheat and $\$150$ per acre of corn. Formulate the problem and find the maximum profit.
▶️ Answer/Explanation
Step 1: Define Variables
Let \( x \) = acres of wheat, \( y \) = acres of corn.
Step 2: Objective Function
Maximize \( P = 200x + 150y \).
Step 3: Constraints
- Total land: \( x + y \le 100 \).
- Labor: \( 3x + 2y \le 240 \).
- \( x, y \ge 0 \).
Step 4: Find Feasible Region & Corner Points
Vertices: (0,0), (0,100), (80,0), (40,60).
Step 5: Evaluate Profit
- (0,100): P=15,000
- (80,0): P=16,000
- (40,60): P=14,000
Optimal Solution: 80 acres wheat, 0 acres corn. Maximum profit = $\$16,000.$