The unequal relationship between two algebraic expressions containing two different variables is represented by linear inequalities in two variables. Hence, the symbols used between the expression in two variables will be ‘<’, ‘>’, ‘≤’ or ‘≥’, but we cannot use equal to ‘=’ symbol here.
The examples of linear inequalities in two variables are:
3x < 2y + 5
8y – 9x > 10
9x ≥ 10/y
x + y ≤ 0
Note: 4x2 + 2x + 5 < 0 is not an illustration of a linear inequality in one variable as x's exponent in the first term is 2. A quadratic inequality exists.
How do you graph inequalities with two variables?
Linear inequalities in two variables have infinite sets or infinitely many ordered pair solutions.
These ordered pairs or the solution sets can be graphed in the appropriate half of a rectangular coordinate plane.
In order to graph inequalities with two variables,
Identify the type of inequality (greater than, less than, greater than or equal to, less than or equal to).
Graph the boundary line - a dashed (in case of strict inequality) or a solid line (in case of non-strict inequalities).
Choose a test point, most probably (0,0) or any other point which is not on the boundary.
Shade the region accordingly. If the test point solves the inequality, shade the region that contains it. Otherwise, shade the opposite side of the boundary line.
Verify with more test points in and out of the region.
Example:
Graph the linear inequality: 2x + 3y > 7
Plot the straight line corresponding to the linear equation: 2x + 3y = 7
Determine any two points (solutions) for this equation: two possible points on the graph can be taken as A (-1, 3), B (2,1) and plot them on the graph.
Determine some specific solutions for the linear inequality 2x + 3y > 7, which can be as follows (2,3), (3,1), (4.5, 0), (0, 3), (1.5, 2).
Plot these five points on the same graph.
All the five points (corresponding to the five solutions) lie above the line.
● Take any point which lies above the line. Its coordinates say (X0, Y0), will satisfy the inequality: 2X0 + 3Y0 > 7
● This means that the solution set for the inequality consists of all points lying above the line.
● Put x = 0, y = 0, which gives 2(0) + 3(0) > 7, which further gives 0 > 7.This doesn't hold true for the given inequality. So, shade the half plane which doesn't include the point (0,0).
Linear Programming is a simple technique where we depict complex relationships through linear functions and then find the optimum points. The important word in the previous sentence is depicted. The real relationships might be much more complex – but we can simplify them to linear relationships.
A linear programming problem can be described completely by linear relationships. It consists of decision variables, which are the unknowns to be determined by the solution to the model, constraints to represent the physical limitation of the system, and an objective function.
Step-by-Step Procedure on solving Linear Programming Problems
Define the Variables
Write the system of inequalities
Graph the System of inequalities
Find the coordinates of the vertices of the feasible region.
Write the function to be maximized or minimized
Substitute the coordinates of the vertices into the functions
Select the greatest or least result, based on your objective function.
Example 1:
A small furniture shop makes tables and chairs which must be processed through assembly and finishing processes. The assembly department is available for 60 hours in every production period while the finishing department is available for 48 hours of work. Making one table requires 4 hours to assemble and 2 hours to finish. Each chair requires 2 hours to assemble and 4 hours to finish. One table contributes PHP 500 to the profit, while a chair contributes PHP 300. Determine the number of tables and chairs to make per production period to maximize profit, assuming all tables and chairs are sold.
Solution:
Step 1. Define the variables
Let x = number of tables
y = number of chairs
Step 2: Write the System of Inequalities
The assembly department is available for 60 hours per production period.
Each table requires 4 hours to assemble and each chair requires 2 hours to assemble. The inequality can be written as 4x + 2y ≤ 60.
Each table requires 2 hours to finish and each chair requires 4 hours to finish. Since the finishing department is available for 48 hours per production period, the inequality can be written as 2x + 4y ≤ 48.
Note that the numbers of tables and chairs cannot be negative, which can be expressed as × ≥ 0 and y ≥ 0. They are called implicit constraints because they are implied in the problem.
The first two inequalities are called explicit constraints because they are explicitly stated in the problem.
Step 3: Graph the System of Inequalities
Step 4: FInd the Coordinates of the vertices of the feasible region.
Solving simultaneously the equations 4x + 2y = 60 and 2x + 4y = 48, the point of intersection is (12, 6). From the graph, a polygon is formed. The vertices of the feasible region are (0, 0), (0, 12), (12, 6), and (15, 0).
Step 5: Write the function to be maximized or minimized.
The function that describes the profit f(x,y) = 500x + 300y. Find the maximum value for this function.
Step 6: Substitute the coordinates of the vertices into the function.
Step 7: Select the greatest or least result.
The maximum value of the function f(x, y) = 500x + 300y is 7,800 at (12, 6). This means that the maximum profit is PHP 7,800 when 12 tables and 6 chairs are produced and sold.
EXERCISE PROBLEM:
A small toy shop makes two wooden toys: wooden cars and trains.
Each wooden car sells for PHP 270, uses PHP 100 worth of raw materials, and takes PHP 14 of labor and overhead costs. Each train sells for PHP 210, uses PHP 90 worth of raw materials, and takes PHP 10 of labor and overhead costs. Each car needs 2 hours finishing and 1 hour carpentry; each train needs 1 hour finishing and 1 hour carpentry. Raw materials are unlimited, but only 100 hours of finishing and 80 hours of carpentry are available each week. Demand for trains is unlimited, but, at most, 40 cars can be sold each week. How many of each toy should be made each week to maximize profits?