Lesson 4: Inequalities and Linear Programming
Estimated time: 45-50 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Graph linear inequalities in two variables and identify solution regions
- Determine whether points satisfy linear inequalities
- Graph systems of linear inequalities and find feasible regions
- Identify vertices (corner points) of feasible regions
- Translate word problems into systems of inequalities with constraints
- Set up objective functions for optimization problems
- Apply the Corner Point Theorem to solve linear programming problems
- Solve real-world optimization problems involving production, diet, transportation, and resource allocation
Introduction: From Equations to Inequalities
In the previous lessons, we solved systems of equations where we found exact points of intersection. Now we shift to inequalities, where solutions are not single points but entire regions in the coordinate plane.
Linear Inequality in Two Variables
A linear inequality in two variables has one of the following forms:
- ax + by < c
- ax + by > c
- ax + by ≤ c
- ax + by ≥ c
where a, b, and c are constants with a and b not both zero.
The solution is a half-plane: all points on one side of the boundary line.
Applications of Linear Inequalities
Linear inequalities and systems of inequalities model constraints in real-world problems:
- Business: Production capacity limits, budget constraints, minimum profit requirements
- Nutrition: Minimum daily requirements for vitamins and nutrients, maximum calorie intake
- Manufacturing: Machine time availability, material supply constraints
- Transportation: Vehicle capacity, route restrictions
- Finance: Investment limits, risk constraints, portfolio diversification
Linear programming is the process of finding the best (optimal) solution to problems with linear constraints and a linear objective function. We will learn systematic methods for solving these optimization problems.
Section 1: Graphing Linear Inequalities
Graphing a linear inequality involves two main steps:
- Graph the boundary line: Replace the inequality symbol with = and graph the line
- Use a solid line for ≤ or ≥ (boundary included in solution)
- Use a dashed line for < or > (boundary not included)
- Determine which side to shade: Use the test point method
- Choose a test point not on the line (often (0,0) if not on the line)
- Substitute the test point into the inequality
- If true, shade the side containing the test point
- If false, shade the opposite side
Example 1: Graph y > 2x + 1
Solution:
Step 1: Graph the boundary line y = 2x + 1
- y-intercept: (0, 1)
- Slope: 2 (rise 2, run 1)
- Use a dashed line because the inequality is > (not ≥)
Step 2: Test point (0, 0)
0 > 2(0) + 1
0 > 1 FALSE
Since the test point (0, 0) does not satisfy the inequality, shade the side of the line that does NOT contain (0, 0).
Answer: Shade the region above the dashed line y = 2x + 1
Example 2: Graph y ≤ -x + 3
Solution:
Step 1: Graph the boundary line y = -x + 3
- y-intercept: (0, 3)
- x-intercept: (3, 0)
- Use a solid line because the inequality is ≤
Step 2: Test point (0, 0)
0 ≤ -(0) + 3
0 ≤ 3 TRUE
Since (0, 0) satisfies the inequality, shade the side containing (0, 0).
Answer: Shade the region below and including the solid line y = -x + 3
Example 3: Graph x < 2
Solution:
This is a vertical line inequality.
Step 1: Graph the boundary line x = 2 (vertical line through (2, 0))
- Use a dashed line because the inequality is <
Step 2: Test point (0, 0)
0 < 2 TRUE
Shade the side containing (0, 0), which is the left side.
Answer: Shade all points to the left of the dashed vertical line x = 2
Example 4: Graph y ≥ -1
Solution:
This is a horizontal line inequality.
Step 1: Graph the boundary line y = -1 (horizontal line through (0, -1))
- Use a solid line because the inequality is ≥
Step 2: Test point (0, 0)
0 ≥ -1 TRUE
Shade the side containing (0, 0), which is above the line.
Answer: Shade all points on and above the horizontal line y = -1
Example 5: Graph 3x - 2y > 6
Solution:
Step 1: Rewrite in slope-intercept form
3x - 2y = 6
-2y = -3x + 6
y = (3/2)x - 3
- y-intercept: (0, -3)
- Slope: 3/2
- Use a dashed line because the inequality is >
Step 2: Test point (0, 0)
3(0) - 2(0) > 6
0 > 6 FALSE
Shade the side NOT containing (0, 0).
Answer: Shade the region below the dashed line y = (3/2)x - 3
When solving for y, if you divide or multiply both sides by a negative number, remember to reverse the inequality symbol.
Example: -2y > -3x + 6 becomes y < (3/2)x - 3 (inequality flips)
Section 2: Systems of Linear Inequalities
A system of linear inequalities consists of two or more inequalities considered simultaneously. The solution is the feasible region: the intersection of all individual solution regions.
Feasible Region
The feasible region (or solution region) is the set of all points that satisfy ALL inequalities in the system simultaneously. It is the overlapping shaded area when all inequalities are graphed.
The vertices (or corner points) are the points where boundary lines intersect at the edges of the feasible region.
Bounded vs. Unbounded Regions
- Bounded region: The feasible region is enclosed and has finite area (like a triangle, rectangle, or polygon)
- Unbounded region: The feasible region extends infinitely in one or more directions
Example 6: System with Bounded Region
Graph the system and identify vertices:
y ≥ x + 1
y ≤ -2x + 7
x ≥ 0
Solution:
Step 1: Graph each inequality
- y ≥ x + 1: Solid line through (0, 1) with slope 1, shade above
- y ≤ -2x + 7: Solid line through (0, 7) with slope -2, shade below
- x ≥ 0: Solid vertical line at x = 0 (y-axis), shade to the right
Step 2: Find the feasible region (where all three regions overlap)
The feasible region is a triangle.
Step 3: Find vertices by solving pairs of boundary equations
| Intersection | Equations | Vertex |
|---|---|---|
| Lines 1 and 3 | y = x + 1 and x = 0 | (0, 1) |
| Lines 2 and 3 | y = -2x + 7 and x = 0 | (0, 7) |
| Lines 1 and 2 | y = x + 1 and y = -2x + 7 | (2, 3) |
For the third vertex:
x + 1 = -2x + 7
3x = 6
x = 2, y = 2 + 1 = 3
Answer: Feasible region is a triangle with vertices (0, 1), (0, 7), and (2, 3)
Example 7: System Forming a Triangle
Graph the system:
x + y ≤ 6
2x + y ≤ 8
x ≥ 0
y ≥ 0
Solution:
Step 1: Rewrite in slope-intercept form and graph
- y ≤ -x + 6: Solid line, slope -1, y-intercept 6, shade below
- y ≤ -2x + 8: Solid line, slope -2, y-intercept 8, shade below
- x ≥ 0: y-axis, shade right
- y ≥ 0: x-axis, shade above
Step 2: Find vertices
- Origin: (0, 0) - intersection of x = 0 and y = 0
- (0, 6) - intersection of y = -x + 6 and x = 0
- (4, 0) - intersection of y = -2x + 8 and y = 0 (solve 0 = -2x + 8)
- (2, 4) - intersection of y = -x + 6 and y = -2x + 8
For the fourth vertex:
-x + 6 = -2x + 8
x = 2, y = -2 + 6 = 4
Answer: Feasible region is a quadrilateral with vertices (0, 0), (0, 6), (2, 4), and (4, 0)
Example 8: System Forming a Rectangle
Graph the system:
x ≥ 1
x ≤ 5
y ≥ 2
y ≤ 6
Solution:
All boundaries are vertical or horizontal lines.
- x = 1 (vertical, solid), shade right
- x = 5 (vertical, solid), shade left
- y = 2 (horizontal, solid), shade above
- y = 6 (horizontal, solid), shade below
Vertices:
(1, 2), (1, 6), (5, 6), (5, 2)
Answer: Feasible region is a rectangle with vertices at the four corners listed above
Example 9: System with Unbounded Region
Graph the system:
x + y ≥ 4
x ≥ 1
y ≥ 0
Solution:
- y ≥ -x + 4: Solid line, slope -1, y-intercept 4, shade above
- x ≥ 1: Vertical solid line at x = 1, shade right
- y ≥ 0: x-axis, shade above
Vertices: The region has two corner points where boundaries meet:
- (1, 3) - intersection of x = 1 and y = -x + 4
- (4, 0) - intersection of y = 0 and y = -x + 4
Answer: Feasible region is unbounded with vertices (1, 3) and (4, 0). The region extends infinitely in the first quadrant.
Example 10: System with No Solution
Graph the system:
y > 2x + 3
y < 2x - 1
Solution:
Both inequalities have boundary lines with the same slope (2), so the lines are parallel.
- y > 2x + 3: Dashed line, shade above
- y < 2x - 1: Dashed line, shade below
The first inequality requires points above y = 2x + 3.
The second inequality requires points below y = 2x - 1.
Since y = 2x + 3 is above y = 2x - 1 for all x, there are no points that can be simultaneously above the first line and below the second line.
Answer: No solution. The feasible region is empty.
Example 11: System with Redundant Inequality
Graph the system:
x + y ≤ 8
x + y ≤ 10
x ≥ 0
y ≥ 0
Solution:
Notice that if x + y ≤ 8 is satisfied, then x + y ≤ 10 is automatically satisfied (since any number ≤ 8 is also ≤ 10).
The second inequality is redundant - it does not further restrict the feasible region.
The effective system is:
x + y ≤ 8
x ≥ 0
y ≥ 0
Vertices: (0, 0), (0, 8), (8, 0)
Answer: Feasible region is a triangle in the first quadrant with vertices (0, 0), (0, 8), and (8, 0). The inequality x + y ≤ 10 is redundant.
Section 3: Linear Programming - Setting Up Problems
Linear programming is a mathematical method for finding the best outcome (maximum profit, minimum cost, etc.) in a model with linear constraints.
Components of a Linear Programming Problem
- Decision variables: The quantities we control (usually x and y)
- Objective function: The quantity we want to maximize or minimize, written as a linear function of the decision variables (e.g., P = 3x + 4y)
- Constraints: Limitations or requirements expressed as linear inequalities (e.g., x + 2y ≤ 40)
- Non-negativity constraints: Usually x ≥ 0 and y ≥ 0 (quantities cannot be negative)
Steps to Set Up a Linear Programming Problem:
- Define decision variables (what are we trying to find?)
- Write the objective function (what are we maximizing or minimizing?)
- Write all constraints as inequalities (what are the limitations?)
- Include non-negativity constraints if applicable
Example 12: Production Problem
Problem: A factory produces two products: widgets and gadgets. Each widget earns a profit of $3, and each gadget earns a profit of $5. The factory has the following constraints:
- Production capacity: Total production cannot exceed 100 units per day
- Labor: Each widget requires 2 hours of labor, each gadget requires 3 hours, and 240 hours of labor are available daily
- Materials: Each widget requires 1 lb of material, each gadget requires 2 lbs, and 150 lbs of material are available daily
Set up the linear programming problem to maximize profit.
Solution:
Step 1: Define variables
Let x = number of widgets produced per day
Let y = number of gadgets produced per day
Step 2: Objective function (maximize profit)
P = 3x + 5y (maximize)
Step 3: Constraints
- Production capacity: x + y ≤ 100
- Labor: 2x + 3y ≤ 240
- Materials: x + 2y ≤ 150
- Non-negativity: x ≥ 0, y ≥ 0
Answer:
Maximize P = 3x + 5y
Subject to:
x + y ≤ 100
2x + 3y ≤ 240
x + 2y ≤ 150
x ≥ 0, y ≥ 0
Example 13: Diet Problem
Problem: A nutritionist is planning a meal using two foods: Food A and Food B. Each serving of Food A costs $2 and provides 3 units of protein and 2 units of fiber. Each serving of Food B costs $3 and provides 2 units of protein and 4 units of fiber. The meal must provide at least 18 units of protein and at least 16 units of fiber. Set up the linear programming problem to minimize cost.
Solution:
Step 1: Define variables
Let x = number of servings of Food A
Let y = number of servings of Food B
Step 2: Objective function (minimize cost)
C = 2x + 3y (minimize)
Step 3: Constraints
- Protein requirement: 3x + 2y ≥ 18
- Fiber requirement: 2x + 4y ≥ 16
- Non-negativity: x ≥ 0, y ≥ 0
Answer:
Minimize C = 2x + 3y
Subject to:
3x + 2y ≥ 18
2x + 4y ≥ 16
x ≥ 0, y ≥ 0
Example 14: Investment Problem
Problem: An investor has $100,000 to invest in two types of bonds: Bond A yields 6% annual return and Bond B yields 8% annual return. Due to risk considerations, the investor wants to invest at most $60,000 in Bond B and at least $20,000 in Bond A. Set up the linear programming problem to maximize annual return.
Solution:
Step 1: Define variables
Let x = amount invested in Bond A (in thousands)
Let y = amount invested in Bond B (in thousands)
Step 2: Objective function (maximize return)
R = 0.06x + 0.08y (maximize, or R = 6x + 8y if using percentages)
Step 3: Constraints
- Total investment: x + y ≤ 100
- Maximum in Bond B: y ≤ 60
- Minimum in Bond A: x ≥ 20
- Non-negativity: x ≥ 0, y ≥ 0 (already satisfied by x ≥ 20)
Answer:
Maximize R = 6x + 8y
Subject to:
x + y ≤ 100
y ≤ 60
x ≥ 20
y ≥ 0
Example 15: Manufacturing Problem with Two Products
Problem: A company manufactures tables and chairs. Each table requires 4 hours of assembly and 2 hours of finishing. Each chair requires 3 hours of assembly and 1 hour of finishing. The company has 240 hours of assembly time and 100 hours of finishing time available per week. Each table generates $60 profit and each chair generates $40 profit. Set up the linear programming problem to maximize weekly profit.
Solution:
Step 1: Define variables
Let x = number of tables produced per week
Let y = number of chairs produced per week
Step 2: Objective function (maximize profit)
P = 60x + 40y (maximize)
Step 3: Constraints
- Assembly time: 4x + 3y ≤ 240
- Finishing time: 2x + y ≤ 100
- Non-negativity: x ≥ 0, y ≥ 0
Answer:
Maximize P = 60x + 40y
Subject to:
4x + 3y ≤ 240
2x + y ≤ 100
x ≥ 0, y ≥ 0
Section 4: Linear Programming - Solving
Corner Point Theorem (Fundamental Theorem of Linear Programming)
If a linear programming problem has an optimal solution, it occurs at a vertex (corner point) of the feasible region.
For bounded feasible regions, both maximum and minimum values always exist and occur at vertices.
For unbounded feasible regions, a maximum may not exist (the objective function can increase without bound).
Procedure for Solving Linear Programming Problems:
- Graph all constraints to find the feasible region
- Identify all vertices (corner points) of the feasible region
- Evaluate the objective function at each vertex
- The vertex that gives the largest value is the maximum; the smallest value is the minimum
- Interpret the result in the context of the problem
Example 16: Maximize Profit with Three Constraints
Problem: Maximize P = 3x + 4y subject to:
x + y ≤ 8
2x + y ≤ 10
x ≥ 0, y ≥ 0
Solution:
Step 1: Graph the feasible region
- y ≤ -x + 8 (shade below)
- y ≤ -2x + 10 (shade below)
- First quadrant
Step 2: Find vertices
- (0, 0) - origin
- (0, 8) - intersection of x = 0 and y = -x + 8
- (5, 0) - intersection of y = 0 and y = -2x + 10
- (2, 6) - intersection of y = -x + 8 and y = -2x + 10
For the fourth vertex, solve:
-x + 8 = -2x + 10
x = 2, y = -2 + 8 = 6
Step 3: Evaluate P = 3x + 4y at each vertex
| Vertex | P = 3x + 4y |
|---|---|
| (0, 0) | P = 3(0) + 4(0) = 0 |
| (0, 8) | P = 3(0) + 4(8) = 32 |
| (5, 0) | P = 3(5) + 4(0) = 15 |
| (2, 6) | P = 3(2) + 4(6) = 6 + 24 = 30 |
Answer: Maximum profit P = 32 occurs at (0, 8)
Example 17: Minimize Cost with Four Constraints
Problem: Minimize C = 2x + 5y subject to:
x + y ≥ 6
2x + y ≥ 10
x ≥ 0
y ≥ 0
Solution:
Step 1: Graph the feasible region
- y ≥ -x + 6 (shade above)
- y ≥ -2x + 10 (shade above)
- First quadrant
Step 2: Find vertices
This is an unbounded region. The corner points are:
- (0, 10) - intersection of x = 0 and y = -2x + 10
- (4, 2) - intersection of y = -x + 6 and y = -2x + 10
- (6, 0) - intersection of y = 0 and y = -x + 6
For the middle vertex, solve:
-x + 6 = -2x + 10
x = 4, y = -4 + 6 = 2
Step 3: Evaluate C = 2x + 5y at each vertex
| Vertex | C = 2x + 5y |
|---|---|
| (0, 10) | C = 2(0) + 5(10) = 50 |
| (4, 2) | C = 2(4) + 5(2) = 8 + 10 = 18 |
| (6, 0) | C = 2(6) + 5(0) = 12 |
Answer: Minimum cost C = 12 occurs at (6, 0)
Interpretation: x = 6, y = 0 minimizes cost at $12
Example 18: Unbounded Feasible Region
Problem: Maximize P = x + 2y subject to:
x + y ≥ 4
x ≥ 0
y ≥ 0
Solution:
Step 1: Graph the feasible region
The feasible region is the area in the first quadrant above the line y = -x + 4. This region is unbounded (extends infinitely).
Step 2: Identify vertices
Vertices: (0, 4) and (4, 0)
Step 3: Evaluate P = x + 2y
| Vertex | P = x + 2y |
|---|---|
| (0, 4) | P = 0 + 2(4) = 8 |
| (4, 0) | P = 4 + 2(0) = 4 |
Step 4: Check if maximum exists
Since the region is unbounded and we can choose arbitrarily large values of x and y while staying in the feasible region, the objective function P = x + 2y can increase without bound.
Answer: No maximum exists. The objective function is unbounded.
Note: The minimum value is P = 4 at vertex (4, 0).
Example 19: Maximize at a Vertex
Problem: Maximize P = 5x + 3y subject to:
x + 2y ≤ 12
3x + y ≤ 15
x ≥ 0, y ≥ 0
Solution:
Step 1: Graph the feasible region
- y ≤ (-1/2)x + 6
- y ≤ -3x + 15
- First quadrant
Step 2: Find vertices
- (0, 0)
- (0, 6) - from x = 0 and x + 2y = 12
- (5, 0) - from y = 0 and 3x + y = 15
- Intersection of x + 2y = 12 and 3x + y = 15
For the fourth vertex:
From x + 2y = 12: x = 12 - 2y
Substitute into 3x + y = 15:
3(12 - 2y) + y = 15
36 - 6y + y = 15
-5y = -21
y = 21/5 = 4.2
x = 12 - 2(4.2) = 12 - 8.4 = 3.6
Vertex: (3.6, 4.2)
Step 3: Evaluate P = 5x + 3y
| Vertex | P = 5x + 3y |
|---|---|
| (0, 0) | P = 0 |
| (0, 6) | P = 0 + 18 = 18 |
| (5, 0) | P = 25 + 0 = 25 |
| (3.6, 4.2) | P = 5(3.6) + 3(4.2) = 18 + 12.6 = 30.6 |
Answer: Maximum profit P = 30.6 occurs at (3.6, 4.2)
Example 20: Complete Word Problem from Setup to Solution
Problem: A farmer has 100 acres of land and wants to plant corn and soybeans. Corn requires 2 hours of labor per acre and generates $300 profit per acre. Soybeans require 3 hours of labor per acre and generate $400 profit per acre. The farmer has 240 hours of labor available. Due to crop rotation requirements, at least 20 acres must be planted with corn. How many acres of each crop should be planted to maximize profit?
Solution:
Step 1: Set up the problem
Let x = acres of corn
Let y = acres of soybeans
Maximize P = 300x + 400y
Subject to:
- Land constraint: x + y ≤ 100
- Labor constraint: 2x + 3y ≤ 240
- Minimum corn: x ≥ 20
- Non-negativity: x ≥ 0, y ≥ 0 (satisfied by x ≥ 20)
Step 2: Graph the feasible region and find vertices
Vertices:
- (20, 0) - intersection of x = 20 and y = 0
- (20, 80) - intersection of x = 20 and x + y = 100
- (60, 40) - intersection of x + y = 100 and 2x + 3y = 240
- (120, 0) would be next, but it violates x + y ≤ 100
Find (60, 40):
From x + y = 100: y = 100 - x
Substitute into 2x + 3y = 240:
2x + 3(100 - x) = 240
2x + 300 - 3x = 240
-x = -60
x = 60, y = 40
Check: Does x = 20, y = 80 satisfy 2x + 3y ≤ 240?
2(20) + 3(80) = 40 + 240 = 280 > 240 NO
So (20, 80) is NOT in the feasible region.
Find intersection of x = 20 and 2x + 3y = 240:
2(20) + 3y = 240
40 + 3y = 240
3y = 200
y = 200/3 ≈ 66.67
Vertex: (20, 66.67)
Find intersection of y = 0 and 2x + 3y = 240:
2x = 240, x = 120
But x + y ≤ 100 means x ≤ 100, so (120, 0) is outside feasible region.
Instead, (100, 0) is a vertex from x + y = 100 and y = 0.
Correct vertices: (20, 0), (100, 0), (60, 40), (20, 66.67)
Step 3: Evaluate P = 300x + 400y
| Vertex | P = 300x + 400y |
|---|---|
| (20, 0) | P = 300(20) + 400(0) = 6000 |
| (100, 0) | P = 300(100) + 400(0) = 30,000 |
| (60, 40) | P = 300(60) + 400(40) = 18,000 + 16,000 = 34,000 |
| (20, 66.67) | P = 300(20) + 400(66.67) = 6000 + 26,668 = 32,668 |
Answer: The farmer should plant 60 acres of corn and 40 acres of soybeans for a maximum profit of $34,000.
Section 5: Real-World Applications
Example 21: Manufacturing - Two Products with Profit Maximization
Problem: A company manufactures laptops and tablets. Each laptop generates $200 profit and each tablet generates $150 profit. Manufacturing constraints are:
- Assembly: Each laptop requires 4 hours, each tablet requires 2 hours. 160 hours available.
- Testing: Each laptop requires 2 hours, each tablet requires 3 hours. 120 hours available.
- Components: At most 50 units total can be produced due to component availability.
How many of each should be manufactured to maximize profit?
Solution:
Let x = number of laptops, y = number of tablets
Maximize P = 200x + 150y
Subject to:
4x + 2y ≤ 160 (assembly)
2x + 3y ≤ 120 (testing)
x + y ≤ 50 (components)
x ≥ 0, y ≥ 0
Vertices of feasible region:
- (0, 0): P = 0
- (0, 40): P = 6000 [from 2x + 3y = 120, x = 0]
- (30, 20): P = 9000 [intersection of 4x + 2y = 160 and 2x + 3y = 120]
- (40, 0): P = 8000 [from 4x + 2y = 160, y = 0]
Find (30, 20): From 4x + 2y = 160, we get y = 80 - 2x
Substitute: 2x + 3(80 - 2x) = 120 → 2x + 240 - 6x = 120 → -4x = -120 → x = 30, y = 20
Answer: Manufacture 30 laptops and 20 tablets for maximum profit of $9,000.
Example 22: Agriculture - Crop Planting with Land and Water Constraints
Problem: A farm has 200 acres of land and 1500 acre-feet of water available for irrigation. Two crops are being considered: wheat and alfalfa. Wheat requires 5 acre-feet of water per acre and generates $400 profit per acre. Alfalfa requires 10 acre-feet of water per acre and generates $600 profit per acre. Find the optimal planting strategy.
Solution:
Let x = acres of wheat, y = acres of alfalfa
Maximize P = 400x + 600y
Subject to:
x + y ≤ 200 (land)
5x + 10y ≤ 1500 (water)
x ≥ 0, y ≥ 0
Simplify water constraint: x + 2y ≤ 300
Vertices:
- (0, 0): P = 0
- (0, 150): P = 90,000 [from x + 2y = 300, x = 0]
- (200, 50): P = 110,000 [intersection of x + y = 200 and x + 2y = 300]
- (200, 0): P = 80,000 [from x + y = 200, y = 0]
Find (200, 50): From x + y = 200, we get x = 200 - y
Substitute: (200 - y) + 2y = 300 → 200 + y = 300 → y = 100
Wait, that gives y = 100, so x = 100, not (200, 50).
Recalculate: x + y = 200 and x + 2y = 300
Subtract: (x + 2y) - (x + y) = 300 - 200 → y = 100
Then x = 200 - 100 = 100
Vertex is (100, 100): P = 400(100) + 600(100) = 40,000 + 60,000 = 100,000
Corrected vertices:
- (0, 0): P = 0
- (0, 150): P = 90,000
- (100, 100): P = 100,000
- (200, 0): P = 80,000
Answer: Plant 100 acres of wheat and 100 acres of alfalfa for maximum profit of $100,000.
Example 23: Transportation - Shipping to Minimize Cost
Problem: A company needs to transport at least 80 tons of goods. They have two types of trucks available: Type A trucks cost $500 per trip and carry 5 tons. Type B trucks cost $700 per trip and carry 8 tons. Due to availability, at most 12 Type A trucks and at most 10 Type B trucks can be used. How many of each truck type should be used to minimize cost?
Solution:
Let x = number of Type A trucks, y = number of Type B trucks
Minimize C = 500x + 700y
Subject to:
5x + 8y ≥ 80 (minimum tonnage)
x ≤ 12 (Type A availability)
y ≤ 10 (Type B availability)
x ≥ 0, y ≥ 0
This is an unbounded region (constraints are ≥ and ≤ mixed). Find corner points where minimum could occur.
Key vertices:
- (16, 0): Not feasible, x > 12
- (0, 10): Check if 5(0) + 8(10) ≥ 80 → 80 ≥ 80 YES. C = 7000
- (12, 2.5): From x = 12 and 5x + 8y = 80 → 60 + 8y = 80 → y = 2.5. C = 500(12) + 700(2.5) = 6000 + 1750 = 7750
- (12, 10): C = 500(12) + 700(10) = 6000 + 7000 = 13,000
Answer: Use 0 Type A trucks and 10 Type B trucks for minimum cost of $7,000.
Example 24: Nutrition - Meal Planning to Minimize Cost
Problem: A dietitian is planning meals using two foods. Food X costs $1.50 per serving and provides 200 calories, 10g protein, and 5g fiber. Food Y costs $2.00 per serving and provides 300 calories, 8g protein, and 12g fiber. The meal plan must provide at least 1200 calories, at least 50g protein, and at least 40g fiber. How many servings of each food minimize cost while meeting requirements?
Solution:
Let x = servings of Food X, y = servings of Food Y
Minimize C = 1.5x + 2y
Subject to:
200x + 300y ≥ 1200 (calories) → Simplify: 2x + 3y ≥ 12
10x + 8y ≥ 50 (protein) → Simplify: 5x + 4y ≥ 25
5x + 12y ≥ 40 (fiber)
x ≥ 0, y ≥ 0
Find vertices of feasible region:
- (0, 10): From x = 0 and 5x + 12y = 40 → y = 40/12 ≈ 3.33. Check: 2(0) + 3(3.33) = 9.99 < 12 NO
- Try (0, 4): 2(0) + 3(4) = 12 YES, 5(0) + 4(4) = 16 < 25 NO
- Intersection of 2x + 3y = 12 and 5x + 4y = 25
Solve 2x + 3y = 12 and 5x + 4y = 25:
From first: x = (12 - 3y)/2
Substitute: 5(12 - 3y)/2 + 4y = 25 → (60 - 15y)/2 + 4y = 25 → 60 - 15y + 8y = 50 → -7y = -10 → y = 10/7 ≈ 1.43
x = (12 - 3(10/7))/2 = (12 - 30/7)/2 = (84/7 - 30/7)/2 = (54/7)/2 = 27/7 ≈ 3.86
Vertices to check:
- (5, 0): C = 7.5 [check: 2(5) + 3(0) = 10 < 12 NO]
- (8, 0): From 5x + 12y = 40, y = 0 → x = 8. C = 12. Check all: 2(8) + 3(0) = 16 ≥ 12 YES, 5(8) + 4(0) = 40 ≥ 25 YES, 5(8) + 12(0) = 40 YES
- (3.86, 1.43): C = 1.5(3.86) + 2(1.43) ≈ 5.79 + 2.86 = 8.65
Answer: Use approximately 3.86 servings of Food X and 1.43 servings of Food Y for minimum cost of approximately $8.65.
Check Your Understanding
Question 1: Which side of the line y = 2x - 3 should be shaded for the inequality y < 2x - 3?
Shade BELOW the dashed line y = 2x - 3. Use test point (0, 0): 0 < 2(0) - 3 → 0 < -3 is FALSE, so shade the side NOT containing (0, 0), which is below the line.
Question 2: Is the point (2, 5) in the solution region of y ≥ x + 2?
Yes. Substitute: 5 ≥ 2 + 2 → 5 ≥ 4 is TRUE. The point (2, 5) satisfies the inequality.
Question 3: Graph the system x + y ≤ 5, x ≥ 1, y ≥ 0 and identify the vertices.
Vertices: (1, 0), (5, 0), (1, 4)
The feasible region is a triangle in the first quadrant bounded by these three points.
Question 4: A company makes products A and B. Product A generates $5 profit, Product B generates $8 profit. Write the objective function to maximize profit.
Let x = number of Product A, y = number of Product B
Objective function: Maximize P = 5x + 8y
Question 5: Each unit of Product A requires 3 hours of labor, each unit of Product B requires 4 hours. 120 hours are available. Write the constraint.
3x + 4y ≤ 120
This represents the labor constraint where total hours used cannot exceed available hours.
Question 6: Given feasible region vertices (0, 0), (4, 0), (3, 2), (0, 5), find the maximum of P = 2x + 3y.
Evaluate at each vertex:
- (0, 0): P = 0
- (4, 0): P = 8
- (3, 2): P = 6 + 6 = 12
- (0, 5): P = 15
Maximum: P = 15 at (0, 5)
Question 7: Why might a linear programming problem have no maximum value?
If the feasible region is unbounded and the objective function increases in the direction that the region extends infinitely, there is no maximum value (the function is unbounded above).
Question 8: A baker makes cookies (profit $2 each) and brownies (profit $3 each). He has 100 oz flour. Cookies need 4 oz, brownies need 5 oz. Set up the problem to maximize profit.
Let x = number of cookies, y = number of brownies
Maximize P = 2x + 3y
Subject to: 4x + 5y ≤ 100, x ≥ 0, y ≥ 0
Key Takeaways
- Linear inequalities in two variables represent half-plane regions; use solid lines for ≤ or ≥, dashed lines for < or >
- Test point method: Choose a point not on the boundary line and test it in the inequality to determine which side to shade
- Systems of inequalities: The solution is the feasible region where all individual solution regions overlap
- Vertices (corner points) are found by solving pairs of boundary equations where lines intersect
- Linear programming components: decision variables, objective function, constraints, non-negativity constraints
- Corner Point Theorem: The optimal solution to a linear programming problem occurs at a vertex of the feasible region
- Solving procedure: Graph feasible region, find vertices, evaluate objective function at each vertex, identify optimal vertex
- Bounded regions always have both maximum and minimum values; unbounded regions may have no maximum (or no minimum)
- Real-world applications include production optimization, resource allocation, diet planning, and cost minimization