Optimization — Mathematics Knowledge System

Overview

Optimization problems ask for the maximum or minimum value of some quantity subject to constraints — the classic 'applied max/min' problems of calculus. The general method: express the quantity to be optimized as a function of a single variable (using constraints to eliminate extra variables), then use derivatives to find and classify its critical points.

Intuition

Whether you're minimizing the material used to build a can of a given volume, or maximizing the area of a fenced rectangular yard with a fixed perimeter, the underlying idea is the same: write down a formula for the quantity you care about, note that its derivative is zero at a peak or valley, and solve for where that happens.

Interactive Graph

Formal Definition

Definition

General optimization procedure for a quantity Q expressed as a function of x on a domain:

Q'(x) = 0 \text{ or } Q'(x) \text{ undefined} \implies \text{candidate extremum at } x

Find critical points of the objective function

\text{Compare } Q(x) \text{ at critical points and domain endpoints (Extreme Value Theorem)}

On a closed interval, the global max/min is among these candidates

Applications

Firms maximize profit by finding the production quantity where marginal revenue equals marginal cost, a direct application of setting the derivative of profit to zero.

Worked Examples

Let x = width (perpendicular to river), y = length (parallel to river). Fencing constraint: 2x + y = 200.
$2x+y=200 \implies y = 200-2x$
Area A(x) = x·y = x(200-2x) = 200x - 2x².
$A(x) = 200x - 2x^2$
Differentiate and set to zero: A'(x) = 200 - 4x = 0 ⟹ x=50.
$A'(x) = 200-4x = 0 \implies x=50$
Check A''(x) = -4 < 0, confirming a maximum. Then y = 200-100 = 100.
$A''(x) = -4 < 0$

Answer: Maximum area is A(50) = 50·100 = 5000 m², with width 50 m and length 100 m.

Practice Problems

Difficulty 5/10

Find two positive numbers whose sum is 20 and whose product is maximized.

Difficulty 6/10

A box with a square base and open top has surface area 300 cm². Maximize its volume.

Difficulty 5/10

A farmer has 400 m of fencing to enclose a rectangular field bordered on one side by a river (no fence needed there). What dimensions maximize the enclosed area?

Difficulty 6/10

A cylindrical can must hold 355 cm³ (a soda can). What radius minimizes the metal used (surface area), and what shape does that give?

Common Mistakes

Common Mistake

Forgetting to reduce the objective function to a single variable before differentiating.

Always use the constraint equation to eliminate one variable first — differentiating a multi-variable expression with respect to only one variable, ignoring the constraint, gives a meaningless result.

Common Mistake

Forgetting to verify whether the critical point is a max or min (or forgetting to check endpoints on a closed domain).

Use the Second Derivative Test or First Derivative Test to classify each critical point, and compare against domain endpoints per the Extreme Value Theorem when the domain is closed and bounded.

Quiz

Summary

Optimization problems maximize or minimize a quantity subject to a constraint, by first reducing it to a single-variable function.
Differentiate the objective function, set the derivative to zero, and solve for critical points.
Classify critical points with the Second Derivative Test (or First Derivative Test) and compare to domain endpoints.
Common problem types: geometric (area/volume/surface area), economic (cost/revenue/profit), and physical (distance/time) optimization.

References

WebsiteWikipedia — Mathematical optimization