diff --git "a/class_7-HandMade Excercise prep EXAM -LP -MathModels - Simplex/answer_1/\342\234\215\357\270\217 Gere a resoluc\314\247a\314\203o completa do exerci\314\201cio, simula.md" "b/class_7-HandMade Excercise prep EXAM -LP -MathModels - Simplex/answer_1/\342\234\215\357\270\217 Gere a resoluc\314\247a\314\203o completa do exerci\314\201cio, simula.md" deleted file mode 100644 index 2596b90..0000000 --- "a/class_7-HandMade Excercise prep EXAM -LP -MathModels - Simplex/answer_1/\342\234\215\357\270\217 Gere a resoluc\314\247a\314\203o completa do exerci\314\201cio, simula.md" +++ /dev/null @@ -1,144 +0,0 @@ - -# 📊 Linear Programming Model — Production Optimization - -## ✅ Problem Statement - -A company, after going through a production streamlining process, ended up with the availability of 3 productive resources: **R1**, **R2**, and **R3**. - -A study on resource usage showed the possibility of producing two products: **P1** and **P2**. After evaluating costs and consulting the sales department, it was found that: - -- **P1 yields a profit of 120 monetary units per unit** -- **P2 yields a profit of 150 monetary units per unit** - -The production department provided the following **resource usage** table: - -| Product | R1/unit | R2/unit | R3/unit | -|---------|---------|---------|---------| -| **P1** | 2 | 3 | 5 | -| **P2** | 4 | 0 | 3 | - -And the **monthly resource availability**: - -| Resource | Monthly Availability | -|----------|----------------------| -| **R1** | 100 | -| **R2** | 90 | -| **R3** | 120 | - ---- - -## 🎯 Objective - -Mathematically model the **Linear Programming (LP)** problem to **maximize profit** under resource constraints. - ---- - -## 🧠 Step-by-Step Modeling - -### 1. 🧮 Decision Variables - -Let: - -x₁ = quantity produced of product P1 -x₂ = quantity produced of product P2 - -Or in LaTeX (for use in documents): - -```latex -x_1 = \text{quantity produced of product P1} \\ -x_2 = \text{quantity produced of product P2} - - - -⸻ - -2. 📈 Objective Function - -Maximize total profit: - -\text{Maximize } Z = 120x_1 + 150x_2 - - - -⸻ - -3. 📏 Resource Constraints - -Each resource has limited availability: - • R1 constraint: - -2x_1 + 4x_2 \leq 100 - - • R2 constraint: - -3x_1 \leq 90 - - • R3 constraint: - -5x_1 + 3x_2 \leq 120 - - - -⸻ - -4. 🚫 Non-Negativity Constraints - -We cannot produce a negative quantity of products: - -x_1 \geq 0, \quad x_2 \geq 0 - - - -⸻ - -🧾 Complete Mathematical Model - -\boxed{ -\begin{cases} -\text{Maximize } Z = 120x_1 + 150x_2 \\ -2x_1 + 4x_2 \leq 100 \\ -3x_1 \leq 90 \\ -5x_1 + 3x_2 \leq 120 \\ -x_1 \geq 0, \quad x_2 \geq 0 -\end{cases} -} - - - -⸻ - -## 📌 Summary Tables - -### 🔢 Profit per Product - -| Product | Profit per Unit (u.m.) | -|:--------|:----------------------:| -| **P1** | 120 | -| **P2** | 150 | - ---- - -### 🧰 Resource Usage per Unit - -| Product | R1/unit | R2/unit | R3/unit | -|:--------|:-------:|:-------:|:-------:| -| **P1** | 2 | 3 | 5 | -| **P2** | 4 | 0 | 3 | - ---- - -### 📦 Monthly Resource Availability - -| Resource | Available Units | -|:---------|:----------------:| -| **R1** | 100 | -| **R2** | 90 | -| **R3** | 120 | - - -⸻ - -🧠 Notes - • This LP model can be solved using methods such as the Simplex Algorithm. - • Can also be implemented in software such as Python (PuLP), MATLAB, or Excel Solver. -