Optimization
Optimization problems ask: What is the maximum or minimum value of something? Derivatives help us find these extreme values by locating where the slope is zero.
Critical Points
Critical Point
A point where f'(x) = 0 or f'(x) is undefined. Maxima and minima can only occur at critical points.
Optimization Strategy
1. Define what you're optimizing (max or min of what?)
2. Write a formula for that quantity
3. Take the derivative and set it equal to zero
4. Solve for critical points
5. Check if it's a max or min (second derivative test)
Box Optimization
Hard
Find the dimensions of a box with square base and volume 32 ft³ that minimizes surface area.
1
Define variables
Let x = side of base, h = height
2
Constraints
Volume: x²h = 32, so h = 32/x²
3
Surface area formula
S = 2x² + 4xh = 2x² + 4x(32/x²) = 2x² + 128/x
4
Differentiate
dS/dx = 4x - 128/x²
5
Set to zero and solve
4x - 128/x² = 0
4x³ = 128
x³ = 32
x = 2∛4 ≈ 3.17 ft
6
Find h
h = 32/x² = 32/(2∛4)² = 2∛4 ft
Answer:
Base ≈ 3.17 ft × 3.17 ft, height ≈ 3.17 ft (a cube!)
Practice
1. Find the maximum value of f(x) = -x² + 4x + 1
f'(x) = -2x + 4 = 0 → x = 2. f(2) = -4 + 8 + 1 = 5
Key Takeaways
To optimize: set derivative = 0, solve for critical points
Use constraints to reduce to one variable
Check endpoints and critical points for max/min
Second derivative test: f''(x) > 0 → minimum, f''(x) < 0 → maximum