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Linear equation problem solving, Lecture notes of Economics

Economics

Linear equation, definition, problems and solutions

Typology: Lecture notes

2021/2022

Uploaded on 11/13/2023

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Aaron Jay De Guzman BSA 2-A
Hesana Mae O. Camba Linear Programming Report
Sheena Marie De Castro
ACTIVITY:
What is Linear Programming?
• Is a model that consists of Linear relationship representing a firm’s decision(s), given an objective and
resource constraints.
• A linear relationship means the interdependence of one variable, called the independent variable and
another variable, called the dependent variable.
• This means a change in the independent variable will affect the dependent variable.
Linear programming also helps the managers to determine solutions (make decisions) for problems to achieve
some objectives (it’s either to maximize profit or to minimize cost) to in which there are restrictions, such as
limited resources or a recipe or perhaps production guidelines.
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Hesana Mae O. Camba Linear Programming Report Sheena Marie De Castro

ACTIVITY:

What is Linear Programming?

  • Is a model that consists of Linear relationship representing a firm’s decision(s), given an objective and resource constraints.
  • A linear relationship means the interdependence of one variable, called the independent variable and another variable, called the dependent variable.
  • This means a change in the independent variable will affect the dependent variable. Linear programming also helps the managers to determine solutions (make decisions) for problems to achieve some objectives (it’s either to maximize profit or to minimize cost) to in which there are restrictions , such as limited resources or a recipe or perhaps production guidelines.

Hesana Mae O. Camba Linear Programming Report Sheena Marie De Castro

Hesana Mae O. Camba Linear Programming Report Sheena Marie De Castro

  • Each is sold at ₱ 100.00, thus profit per unit of pizza produced and sold would be ₱ 25.00.
  • Meanwhile, the cost per burger is computed as follows:
  • Each burger is sold at ₱ 75.00, thus the profit per burger is ₱ 21.00.
  • With the given variables of: X = the number of pizza to produce and sell Y = the number of burgers to produce and sell Our objective function is represented as follows: Step 3: Define the Constraints
  • Constraints are the restrictions or limitations in achieving our objective that is to maximize profit and minimize costs.
  • In the activity, four are used for production – the dough, meat, veggies, and cheese – all of which are limited.
  • Production of pizza and burger requires all four resources.

Hesana Mae O. Camba Linear Programming Report Sheena Marie De Castro

  • How many resources do we have?
    • We have four resources.
  • Each resource is limited, and we have to utilize most of them to maximize our profit.
  • This can be represented as follows:

Hesana Mae O. Camba Linear Programming Report Sheena Marie De Castro The values are false because the total number of dough that should be used must not exceed the total number of dough available.

Solutions:

Under this method, constraint equations are solved simultaneously at the optimal extreme point to determine the variable solution values.

  • To solve the problem, two constraints will be used. Dough: 2x + 1y ≤ 23 Meat: 3x + 1y ≤ 30
  • Convert the inequality sign to an equal sign. Dough: 2x + 1y = 23 Meat: 3x + 1y = 30
  • Compute either x or y for both equations. 2x + 1y = 23 y = - 2x + 3x + 1y = 30 y = - 3x + 30
    • 2x + 23 = - 3x + 30
    • 2x + 3x = 30 - 23

x = 7 ——— Number of pizzas to be produced and sold.

  • Using the same value, solve for y. Dough: 2x + 1y = 23 2(7) + y = 23 14 + y = 23 y = 23 - 14

y = 9 ——— Number of burgers to be produced and sold.

Hesana Mae O. Camba Linear Programming Report

Sheena Marie De Castro

  • To determine whether the values of x and y are correct, test the values with the

    • Veggies: 1x + 2y ≤ • Use the two other constraints equations.
    • Cheese: 2x + 3y ≤
    • 1x + 2y =
    • 1x = - 2y +
    • x = - 2y +
    • 2x + 3y =
    • 2x = - 3y +
    • x = - 3/2y + 41/
      • 2y + 25 = - 3/2y + 41/
      • 4y + 50 = - 3y +
      • 4y + 3y = 41 -
      • y = - 9 or y =

  • Dough: 2x + 1y ≤ constraints equations.

  • Meat: 3x + 1y ≤

  • Veggies: 1x + 2y ≤

  • Cheese: 2x + 3y ≤

  • Final Restriction: x ≥ 0, y ≥

  • 2x + 1y ≤ Dough:

  • 2(7) + 1(9) ≤

  • 14 + 9 ≤

  • 23 ≤

  • 3x + 1y ≤ Meat:

  • 3(7) + 1(9) ≤

  • 21 + 9 ≤

  • 30 ≤

  • 1x + 2y ≤ Veggies:

  • 1(7) + 2(9) ≤

  • 7 + 18 ≤

  • 25 ≤