The core idea
The derivative f′(x) is the limit of [f(x+h)−f(x)]/h as h approaches zero, when that limit exists. Optimisation combines a meaningful objective, a feasible domain, derivative information and boundary checks; solving f′(x) = 0 alone is insufficient.
1. Bring two inputs closer without dividing by zero
For a nonzero increment h, [f(x+h)−f(x)]/h is an average rate over a small interval. On a graph it is the slope of a secant through two points. To define the derivative, examine what this quotient approaches as h becomes arbitrarily close to zero from either side. We do not substitute h = 0 into the original fraction. For f(x) = x², expansion gives [(x+h)²−x²]/h = (2xh+h²)/h = 2x+h for h ≠ 0. As h approaches zero, this tends to 2x, so f′(x) = 2x. At x = 2, increments 0.1 and −0.1 give slopes 4.1 and 3.9; both approach 4 as the interval shrinks. The algebra, rather than a few numerical trials alone, establishes the limiting value.
Sources: OpenStax: Defining the Derivative ↗
2. Rules summarise limits; units preserve meaning
For positive integer n, the power rule gives the derivative of xⁿ as nxⁿ⁻¹. A constant has derivative zero. Derivatives distribute over sums, and constant multipliers remain: the derivative of 4x³−2x+7 is 12x²−2. These rules follow from limits and save repeating the expansion each time. The derivative is another function, so evaluate it at the required input after differentiating. Its units are output units divided by input units: metres per second for position versus time, for example. It is not generally f(x)/x. For f(x) = x² at x = 2, the tangent line is y = 4+4(x−2). Near x = 2 this line approximates the curve; farther away its error grows because the square also contains a quadratic increment term.
Sources: OpenStax: Differentiation Rules ↗ · OpenStax: Defining the Derivative ↗
3. Worked case: average and instantaneous velocity
Consider the invented position model s(t) = 3t²+2t metres during 0 ≤ t ≤ 4 seconds. Between t = 2 and t = 3, the positions are 16 m and 33 m, so average velocity is (33−16)/(3−2) = 17 m/s. Differentiating gives v(t) = s′(t) = 6t+2 m/s, so instantaneous velocity is 14 m/s at two seconds and 20 m/s at three seconds. These values bracket the average because velocity is increasing throughout the interval. Differentiating once more gives acceleration 6 m/s². A position value, a velocity and an acceleration answer different questions and carry different units. The model makes velocity linear and acceleration constant; those are properties of this chosen formula, not claims about every moving object or about unobserved times beyond its stated interval.
Sources: OpenStax: Defining the Derivative ↗ · OpenStax: Differentiation Rules ↗
4. Turn a design question into a feasible objective
Optimisation asks for the greatest or least value that satisfies the constraints. First identify what is being maximised, choose variables with units, and use the constraints to express the objective in one variable. Then state its feasible domain. A continuous function on a closed bounded interval attains a maximum and minimum. To find them, compare values at endpoints and at interior critical points, where the derivative is zero or fails to exist. A zero derivative identifies a horizontal tangent, not automatically a maximum. For x³, the derivative is zero at x = 0, but the function keeps increasing through that point. The function |x| has a minimum at zero even though its left and right slopes disagree there. The original objective and domain decide which candidate is best.
Sources: OpenStax: Applied Optimization Problems ↗ · OpenStax: Defining the Derivative ↗ · OpenStax: Maxima and Minima ↗
5. Worked case: maximise a model box’s volume
A rectangular sheet measures 30 cm by 20 cm. Imagine removing equal squares of side x from the four corners and folding the sides into an open box, ignoring material thickness. Its height is x, length 30−2x and width 20−2x, so V(x) = x(30−2x)(20−2x) = 600x−100x²+4x³ cm³. A nondegenerate box requires 0 < x < 10; the limiting endpoints have volume zero. Differentiate: V′(x) = 600−200x+12x². Setting this to zero gives 3x²−50x+150 = 0, with roots (25 ± 5√7)/3. Only x ≈ 3.92375 cm is feasible; the other root is approximately 12.74292 cm. At the feasible candidate, dimensions are about 22.1525 cm by 12.1525 cm by 3.92375 cm, giving volume approximately 1,056.31 cm³.
Constrain the design before maximising
| Cut x (cm) | Volume (cm³) |
|---|---|
| 0 (limit) | 0 |
| 2 | 832 |
| 3.92375… | 1056.31… |
| 6 | 864 |
| 10 (limit) | 0 |
Sources: OpenStax: Applied Optimization Problems ↗ · OpenStax: Quadratic Equations ↗
6. Prove the maximum and interpret its precision
The box calculation is unfinished until the candidate is checked. V′ is positive between 0 and 3.92375 and negative from there to 10, so volume increases and then decreases throughout the feasible interval. The continuous extension has V(0) = V(10) = 0. Together these facts establish the global maximum, not merely a stationary point. As a numerical check, x = 2 gives 832 cm³ and x = 6 gives 864 cm³, both smaller. A physical sheet has thickness and folds, and cutting has limited precision; the many decimal places belong to the ideal model. Rounding a design should be followed by recalculating its volume. Mathematical optimisation gives the best value under the chosen objective and assumptions. It does not automatically optimise strength, ease of construction or material waste as additional goals.
PUT IT INTO PRACTICE
Practice: optimise three fenced sides
- A model rectangular enclosure uses an existing straight wall for one side and 40 m of fencing for the other three. Let each side perpendicular to the wall be x and the parallel fenced side be y.
- Use 2x+y = 40 to express area A in terms of x. State the interval in which both lengths are nonnegative.
- Differentiate, find an interior stationary point and compare its area with the limiting endpoint areas.
- Solution: y = 40−2x and A(x) = 40x−2x² for 0 ≤ x ≤ 20. A′ = 40−4x is zero at x = 10; then y = 20 and area is 200 m². The derivative changes from positive to negative, and both endpoint areas are zero. This verifies the maximum for this three-sided fencing model.
Check your understanding
Why not put h = 0 into the difference quotient?
That creates division by zero. The derivative is a limit of values for nonzero h, not the value of that original fraction at zero.
Is f′(2) the same as f(2)/2?
Generally no. A derivative measures local change. For f(x) = x², these values are 4 and 2 respectively.
Why does position differentiation produce metres per second?
The difference quotient divides a position change in metres by a time change in seconds; the limiting process preserves those units.
Does every zero derivative mark an extreme?
No. The function x³ has derivative zero at the origin but no maximum or minimum there. Check nearby behaviour and the domain.
Why reject the box root near 12.74 cm?
It makes 20−2x negative, so the supposed box width is impossible. Algebraic candidates must satisfy the original geometric constraints.
Does maximum volume imply the best box for every purpose?
No. Strength, durability, cost and construction constraints are different objectives or restrictions absent from this idealised model.