MathIsimple
Lesson 6.3: Real-World Applications & Modeling

Master Real-World Applications & Modeling

Apply mathematics to real life! Learn to model complex scenarios from science, engineering, and everyday situations using mathematical concepts and problem-solving strategies.

Learning Objectives

Translate real-world problems into mathematical models
Apply multiple mathematical concepts to solve problems
Validate solutions in real-world context
Communicate mathematical reasoning clearly

Engineering Applications

Example 1: Bridge Design

Problem: An engineer needs to design a bridge that can support 50,000 kg. The bridge will be 100 meters long with support pillars every 25 meters. Each pillar must support an equal share of the weight.

Step 1: Determine number of support pillars

Bridge length: 100m, Pillar spacing: 25m

Number of pillars = 100 ÷ 25 = 4 pillars

Step 2: Calculate weight per pillar

Total weight: 50,000 kg

Weight per pillar = 50,000 ÷ 4 = 12,500 kg

Step 3: Apply safety factor

Safety factor: 2.5 (standard for bridges)

Required capacity = 12,500 × 2.5 = 31,250 kg per pillar

Scientific Applications

Example 2: Population Growth

Problem: A bacteria colony starts with 1000 cells and doubles every 3 hours. How many cells will there be after 12 hours? When will the population reach 1 million cells?

Step 1: Model the growth

Initial population: 1000 cells

Doubling time: 3 hours

Growth rate: 2^(t/3) where t is time in hours

Step 2: Calculate population after 12 hours

Population = 1000 × 2^(12/3) = 1000 × 2^4 = 1000 × 16 = 16,000 cells

Step 3: Find time for 1 million cells

1,000,000 = 1000 × 2^(t/3)

1000 = 2^(t/3)

t/3 = log₂(1000) ≈ 9.97

t ≈ 29.9 hours

Economic Applications

Example 3: Business Optimization

Problem: A company produces widgets. The cost function is C(x) = 0.5x² + 10x + 100, and the revenue function is R(x) = 25x, where x is the number of widgets. Find the profit-maximizing production level.

Step 1: Find the profit function

P(x) = R(x) - C(x) = 25x - (0.5x² + 10x + 100)

P(x) = -0.5x² + 15x - 100

Step 2: Find the vertex (maximum point)

For P(x) = ax² + bx + c, vertex at x = -b/(2a)

x = -15/(2 × (-0.5)) = -15/(-1) = 15

Step 3: Calculate maximum profit

P(15) = -0.5(15)² + 15(15) - 100

P(15) = -112.5 + 225 - 100 = 12.5

Maximum profit: $12.50 at 15 widgets

Environmental Applications

Example 4: Carbon Footprint Analysis

Problem: A family wants to reduce their carbon footprint. They currently drive 15,000 miles/year in a car that gets 25 mpg. Gasoline produces 19.6 lbs CO₂ per gallon. If they switch to a hybrid that gets 50 mpg, how much CO₂ will they save annually?

Step 1: Calculate current CO₂ emissions

Current gallons used = 15,000 ÷ 25 = 600 gallons/year

Current CO₂ = 600 × 19.6 = 11,760 lbs/year

Step 2: Calculate new CO₂ emissions

New gallons used = 15,000 ÷ 50 = 300 gallons/year

New CO₂ = 300 × 19.6 = 5,880 lbs/year

Step 3: Calculate savings

CO₂ saved = 11,760 - 5,880 = 5,880 lbs/year

Percentage reduction = (5,880 ÷ 11,760) × 100% = 50%

Advanced Modeling Techniques

Optimization Models

Linear Programming:

Maximize/minimize objective function

Use: Resource allocation and production planning

Simulation Models

Monte Carlo:

Random sampling for complex systems

Use: Risk analysis and system behavior prediction

Real-world Applications

Environmental Science

Climate Modeling

Temperature and precipitation predictions

Population dynamics and ecosystem analysis

Economics & Finance

Market Analysis

Price prediction and trend analysis

Portfolio optimization and risk management

Social Sciences

Urban Planning

Traffic flow and city development

Social network analysis and behavior modeling