Optimization Problems
Optimization problems ask you to maximize or minimize some quantity subject to a constraint, using critical points of a single-variable function built from the constraint.
The optimization recipe
1) Identify the quantity to optimize (write a formula for it, possibly in two variables). 2) Use the given constraint to eliminate one variable, leaving a single-variable function. 3) Find critical points of that function using the first derivative. 4) Justify the max/min with the first or second derivative test (or by checking endpoints, if the domain is a closed interval).
Don't forget domain restrictions from the physical context (e.g., lengths and areas can't be negative) — these often turn an unbounded critical point search into a closed-interval problem where endpoints matter too.
Worked Example
A rectangular field with 100 m of fencing (used to enclose 3 sides against an existing wall) — maximize the enclosed area.
- Let x = width (two sides perpendicular to the wall), y = length (parallel to wall). Fencing constraint: 2x + y = 100, so y = 100-2x.
- Area: A(x) = x·y = x(100-2x) = 100x - 2x².
- A'(x) = 100 - 4x. Set to 0: x=25.
- A''(x) = -4 < 0, confirming a maximum. Then y = 100-2(25) = 50.
Answer: Maximum area is 25×50 = 1250 m², achieved at x=25, y=50.
Key terms (1)
- Constraint
- A given relationship between variables (e.g., fixed perimeter) used to reduce the optimization problem to one variable.
Practice Quiz
In an optimization problem, why do you substitute the constraint equation into the quantity you're optimizing?