Discover how quadratic functions model real-world phenomena! Learn to solve optimization problems, analyze projectile motion, and calculate areas using the power of quadratics.
• Step 1: Define variables and write the function
• Step 2: Find the vertex (maximum or minimum point)
• Step 3: Interpret the result in context
• Step 4: Check if the solution makes sense
Problem: A farmer has 200 meters of fencing to enclose a rectangular area. What dimensions will give the maximum area?
Step 1: Define variables
Let x = width, y = length
Perimeter: 2x + 2y = 200
Therefore: y = 100 - x
Step 2: Write area function
Area = x × y = x(100 - x) = 100x - x²
A(x) = -x² + 100x
Step 3: Find vertex (maximum)
x = -b/(2a) = -100/(2×(-1)) = 50
y = 100 - 50 = 50
Maximum area = 50 × 50 = 2,500 m²
Result: A square with sides of 50 meters gives maximum area of 2,500 m²
Problem: A company sells widgets for $20 each. The cost to produce x widgets is C(x) = 0.5x² + 10x + 100. How many widgets should be produced to maximize profit?
Step 1: Write profit function
Revenue = 20x
Cost = 0.5x² + 10x + 100
Profit = Revenue - Cost
P(x) = 20x - (0.5x² + 10x + 100)
P(x) = -0.5x² + 10x - 100
Step 2: Find vertex (maximum)
x = -b/(2a) = -10/(2×(-0.5)) = 10
P(10) = -0.5(10)² + 10(10) - 100 = -50 + 100 - 100 = -50
Analysis: The company loses money at all production levels!
This suggests the selling price is too low or costs are too high.
h(t) = -½gt² + v₀t + h₀
• h(t): height at time t
• g: acceleration due to gravity (≈ 32 ft/s² or 9.8 m/s²)
• v₀: initial velocity
• h₀: initial height
Problem: A ball is thrown upward from a height of 5 feet with an initial velocity of 48 ft/s. When will it hit the ground? What is the maximum height?
Step 1: Write the height function
h(t) = -½(32)t² + 48t + 5
h(t) = -16t² + 48t + 5
Step 2: Find maximum height (vertex)
t = -b/(2a) = -48/(2×(-16)) = 1.5 seconds
h(1.5) = -16(1.5)² + 48(1.5) + 5 = -36 + 72 + 5 = 41 feet
Step 3: Find when ball hits ground (h = 0)
0 = -16t² + 48t + 5
Using quadratic formula: t = 3.1 seconds (positive root)
Results:
• Maximum height: 41 feet at 1.5 seconds
• Ball hits ground at 3.1 seconds
Problem: A rectangular garden is to be built against a wall. If 60 feet of fencing is available for the three open sides, what dimensions will maximize the garden area?
Step 1: Set up the problem
Let x = width (perpendicular to wall)
Let y = length (parallel to wall)
Fencing: 2x + y = 60, so y = 60 - 2x
Step 2: Write area function
Area = x × y = x(60 - 2x) = 60x - 2x²
A(x) = -2x² + 60x
Step 3: Find maximum
x = -b/(2a) = -60/(2×(-2)) = 15 feet
y = 60 - 2(15) = 30 feet
Maximum area = 15 × 30 = 450 ft²
Result: Width = 15 feet, Length = 30 feet, Area = 450 ft²
Problem: A window frame has a rectangular opening with a semicircular top. If the perimeter of the entire frame is 20 feet, what dimensions will maximize the light area?
Step 1: Define variables
Let x = width of rectangle
Let y = height of rectangle
Perimeter: 2y + x + ½πx = 20
Therefore: y = 10 - x/2 - πx/4
Step 2: Write area function
Rectangle area = xy
Semicircle area = ½π(x/2)² = πx²/8
Total area = xy + πx²/8
Step 3: Substitute and simplify
A(x) = x(10 - x/2 - πx/4) + πx²/8
A(x) = 10x - x²/2 - πx²/4 + πx²/8
A(x) = 10x - x²/2 - πx²/8
Step 4: Find maximum (using calculus or vertex formula)
This requires more advanced techniques, but the concept is the same!
Identify what you're trying to find and what information is given
Choose letters to represent unknown quantities
Express the quantity to be optimized as a function of one variable
Use x = -b/(2a) to find the maximum or minimum point
Answer the original question in context
Always verify that your answer is reasonable in the context of the problem
Include appropriate units (feet, meters, seconds, etc.) in your final answer
Some values might not make sense (negative dimensions, negative time, etc.)
Problem 1:
A ball is thrown upward from ground level with initial velocity 64 ft/s. Find the maximum height and time to hit the ground.
h(t) = -16t² + 64t
Maximum height: 64 feet at 2 seconds
Hits ground at 4 seconds
Problem 2:
A rectangular field is to be fenced with 300 meters of wire. What dimensions will maximize the area?
Let x = width, y = length
2x + 2y = 300, so y = 150 - x
Area = x(150 - x) = -x² + 150x
Maximum at x = 75 meters
Dimensions: 75m × 75m (square)
Problem 3:
A company's profit is P(x) = -2x² + 100x - 800, where x is the number of units sold. Find the maximum profit.
Maximum at x = 25 units
P(25) = -2(25)² + 100(25) - 800 = 450
Maximum profit: $450