Apply your knowledge of linear functions to solve complex real-world problems in business, science, and everyday life scenarios.
Using linear functions to optimize business decisions
Scenario: A coffee shop sells coffee for $3.50 per cup. Fixed costs are $200 per day, and variable costs are $1.50 per cup.
Linear Functions:
• Revenue: R(x) = 3.50x
• Cost: C(x) = 1.50x + 200
• Profit: P(x) = R(x) - C(x) = 2x - 200
Break-even Analysis:
Break-even when P(x) = 0: 2x - 200 = 0 → x = 100 cups
Scenario: A company finds that for every $1 increase in price, they sell 10 fewer units. At $20, they sell 100 units.
Demand Function:
q = -10p + 300 (where q = quantity, p = price)
Revenue Function:
R(p) = p × q = p(-10p + 300) = -10p² + 300p
Analyzing movement using linear functions
Scenario: Car A starts 200 miles away and drives at 60 mph. Car B starts at the same time and drives at 40 mph toward Car A.
Distance Functions:
• Car A: d₁(t) = 200 - 60t
• Car B: d₂(t) = 40t
Meeting Time:
When d₁(t) = d₂(t): 200 - 60t = 40t
200 = 100t → t = 2 hours
Scenario: A ball is thrown upward with initial velocity 30 m/s from a height of 2 meters.
Height Function:
h(t) = -4.9t² + 30t + 2
Maximum Height:
At vertex: t = -b/(2a) = -30/(-9.8) ≈ 3.06 seconds
h(3.06) ≈ 47.9 meters
Making informed decisions using linear functions
Plan A: $30/month + $0.10 per minute
Plan B: $50/month + $0.05 per minute
Question: Which plan is better for different usage levels?
Cost Functions:
• Plan A: C₁(m) = 30 + 0.10m
• Plan B: C₂(m) = 50 + 0.05m
Break-even Point:
30 + 0.10m = 50 + 0.05m
0.05m = 20 → m = 400 minutes
Recommendation:
• < 400 minutes: Choose Plan A
• > 400 minutes: Choose Plan B
• = 400 minutes: Both plans cost the same
Finding the best solution using linear functions
Scenario: You have 100 feet of fencing to enclose a rectangular area. One side is against a wall, so you only need to fence 3 sides.
Find the dimensions that maximize the area.
Solution:
Let x = width, y = length
Constraint: 2x + y = 100 → y = 100 - 2x
Area: A(x) = x(100 - 2x) = 100x - 2x²
Maximum at x = 25 feet, y = 50 feet
Maximum area: 1,250 square feet
Scenario: A factory can produce Product A (profit $5) or Product B (profit $8). Each unit of A takes 2 hours, each unit of B takes 3 hours. Total available time: 120 hours.
How many of each should be produced to maximize profit?
Solution:
Let x = units of A, y = units of B
Constraint: 2x + 3y ≤ 120
Profit: P(x,y) = 5x + 8y
Optimal solution: x = 0, y = 40
Maximum profit: $320
Explore more sophisticated uses of linear functions in real-world scenarios
Definition: Optimizing a linear objective function
Subject to linear constraints (inequalities)
Example: Maximize profit
P = 3x + 2y (profit function)
Subject to: x + y ≤ 10, x ≥ 0, y ≥ 0
Applications:
• Resource allocation
• Production planning
• Transportation optimization
Purpose: Find the best-fit line through data points
Minimize the sum of squared errors
Formula: y = mx + b
Where m and b are calculated from data
Applications:
• Sales forecasting
• Price prediction
• Trend analysis
Learn the systematic approach to solving real-world problems with mathematics
Read the problem
Identify what's given and what's asked
Identify variables
Define what each variable represents
Write equations
Translate words into mathematical expressions
Set up system
Create equations based on relationships
Choose method
Substitution, elimination, or graphing
Calculate answer
Show all work step by step
Check solution
Substitute back into original equations
Interpret result
Answer in context of the problem
Congratulations! You've mastered the fundamentals of linear functions and systems