Math Problem Statement

III. Carry out the Plan. IV. Look back C. CHECK C.1 Check for Understanding Solve the given Linear Programming maximization problem. Also, just a heads up, this activity is going to count for 20% of your P3 Exam grade. So, make sure you're putting your best effort into it and really engage with your partner. Let's make it count and let's get started! Imagine there's a local barangay cafe that offers two types of services: Coffee and Snacks. Each service requires the assistance of two staff members, Barista A and Barista B. Each Coffee service takes 1 hour of assistance by Barista A and 2 hours by Barista B, while each Snacks service takes 3 hours by Barista A and 1 hour by Barista B. The cafe has Barista A available for 25 hours each week and Barista B available for 30 hours each week. The profit per Coffee service is ₱80, and the profit per Snacks service is ₱120. How many services of each type the cafe should schedule each week to maximize profits?

Solution

To solve this Linear Programming (LP) problem, we’ll need to:

  1. Define the variables.
  2. Set up the objective function to maximize.
  3. Write the constraints based on the hours available for each barista.

Step 1: Define the Variables

Let:

  • xx = number of Coffee services provided each week.
  • yy = number of Snacks services provided each week.

Step 2: Objective Function

The goal is to maximize the weekly profit.

Profit per Coffee service = ₱80
Profit per Snacks service = ₱120

The objective function, representing total profit PP, is: P=80x+120yP = 80x + 120y

Step 3: Constraints

The cafe has limited hours for each barista, leading to constraints based on their weekly availability.

Barista A's Constraint

Each Coffee service requires 1 hour of Barista A’s time, and each Snacks service requires 3 hours. Since Barista A is available for a maximum of 25 hours weekly: 1x+3y251x + 3y \leq 25

Barista B's Constraint

Each Coffee service requires 2 hours of Barista B’s time, and each Snacks service requires 1 hour. Since Barista B is available for a maximum of 30 hours weekly: 2x+1y302x + 1y \leq 30

Non-negativity Constraints

The number of services cannot be negative: x0andy0x \geq 0 \quad \text{and} \quad y \geq 0

Step 4: Solve the System

The problem now is: Maximize: P=80x+120yP = 80x + 120y Subject to: 1x+3y251x + 3y \leq 25 2x+1y302x + 1y \leq 30 x0,y0x \geq 0, \quad y \geq 0

Step 5: Graphical Solution

To find the optimal solution, we can solve this graphically by plotting each constraint and finding the feasible region. We then evaluate the profit function P=80x+120yP = 80x + 120y at each vertex of the feasible region to determine which combination of xx and yy maximizes PP.

Let's go over these calculations to get the exact answer. If you'd like further details on solving it graphically or using another method, just let me know!


Would you like a step-by-step calculation for each constraint? Here are related questions to explore further:

  1. How would you solve a linear programming problem with more than two variables?
  2. What is the importance of defining non-negativity constraints in LP problems?
  3. How can duality theory be applied to this maximization problem?
  4. What methods other than graphical analysis can solve this LP problem?
  5. How does changing the profit per service impact the optimal solution?

Tip: Always check each constraint to ensure that the chosen values for xx and yy satisfy the feasible region for the problem.

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Maximization Problem
Constraints
Objective Function

Formulas

Objective function: P = 80x + 120y
Constraint for Barista A: 1x + 3y ≤ 25
Constraint for Barista B: 2x + 1y ≤ 30
Non-negativity constraints: x ≥ 0, y ≥ 0

Theorems

Feasible Region Theorem

Suitable Grade Level

Grades 10-12