Optimization (Calculus)
In short: Optimization applies derivatives to locate the input giving a best outcome, the biggest area, the lowest cost or the quickest trip, when resources are limited.
AP Calculus ABCalculusAP Calculus AB Unit 5
What is Optimization?
Optimization converts a real situation with a cap, such as a fixed budget or a fixed length of material, into a formula in a single letter and then hunts for its best value. Best means largest area, smallest cost or shortest time.
The idea is that a smooth function usually reaches its maximum or minimum where the tangent line is flat, so you solve \(f^{\prime}(x) = 0\), then confirm the answer is truly a maximum or minimum, and finally compare with the endpoints of the allowed interval. The word problem supplies the function and the limits on \(x\).
How to spot it
- A word problem asks for the largest, smallest, cheapest, fastest or greatest amount under a stated limit.
- A fixed quantity such as 84 feet of fencing or 300 square inches of cardboard constrains the shape.
- Two variables are linked by a condition, and you can eliminate one to write a single-variable function.
- The answer must be a dimension, a price or a time, not just the maximum value.
The method
- Draw the situation, name the variables, and write the quantity to optimize as a formula.
- Use the constraint to rewrite that formula with one variable, and state the allowed interval for it.
- Differentiate, solve \(f^{\prime}(x) = 0\), and test the critical numbers and the endpoints.
- Answer the question in words with units, giving every dimension the problem asked for.
Optimization: solved example
Problem
A school garden club has 84 feet of fencing and will enclose a rectangular plot against a long brick wall, so only three sides need fence. What dimensions give the largest area?
Solution
Let \(x\) be the length of each side perpendicular to the wall. The side parallel to the wall is \(84 – 2x\), so \(A(x) = x(84 – 2x) = 84x – 2x^2\) with \(0 \le x \le 42\).
\(A^{\prime}(x) = 84 – 4x = 0\) gives \(x = 21\). Since \(A^{\prime\prime} = -4 < 0\), it is a maximum; the endpoints give area 0.
The parallel side is \(84 – 42 = 42\). The best plot is 21 feet by 42 feet, area 882 square feet.
Try it
Question 1
Two numbers add up to 40. What is the largest possible product?
Show the answer
With numbers \(x\) and \(40 – x\), \(P = 40x – x^2\), \(P^{\prime} = 40 – 2x = 0\) at \(x = 20\). Both numbers are 20 and the product is 400.
Question 2
A baker folds an open tray from an 18-inch square of foil by trimming a small square of side \(x\) off each corner. Which \(x\) gives the roomiest tray?
Show the answer
\(V = x(18 – 2x)^2\), so \(V^{\prime} = (18 – 2x)(18 – 6x) = 0\) gives \(x = 3\) (since \(x = 9\) gives volume 0). \(V(3) = 3 \cdot 144 = 432\) cubic inches.
Optimization vs. Extreme Value Theorem
The theorem only promises that a maximum and minimum exist on a closed interval. Optimization is the full job: build the function from the story, locate the best input, justify that it is the best, and report it with units.
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