Function analysis and optimisation

Questions come before derivatives

An isolated formula does not determine an optimisation problem. The expression A(x)=x(20−x), for example, can be studied over all real numbers as a polynomial; if x and 20−x are lengths, only 0<x<20 represents nondegenerate rectangles. The contextual domain may be smaller than the algebraic domain. We must also decide what to maximise: area, efficiency, revenue or another quantity.

Start by naming variables with units, drawing the configuration and writing the constraints. Then eliminate variables, where possible, to obtain a function of one input. An optimal answer requires existence and admissibility: a value that makes the derivative zero may lie outside the permitted interval or violate an original condition. The drawing organises relationships, but constraints must appear explicitly in equations and inequalities.

The function's sign and the derivative's sign

The sign of f(x) indicates whether the graph is above or below the horizontal axis. The sign of f′(x) describes variation: if f′ is positive on an interval, f is increasing there; if negative, f is decreasing. These are different questions. The function −10+x increases on (0,2), despite taking only negative values on that interval.

To construct a variation table, find where the derivative is zero or ceases to exist and split the domain at those points and its gaps. Determine the sign on each interval through factoring or another argument. A single test value works when we already know there are no zeros or changes of definition on the interval. Then translate signs into increasing and decreasing arrows and calculate outputs at relevant points.

Local extrema, absolute extrema and candidates

A local maximum exceeds sufficiently nearby outputs; an absolute maximum exceeds every output in the domain under consideration. A minimum follows the reverse comparison. If there is an extremum at an interior point where the derivative exists, it must be zero: a strictly positive or negative rate would allow greater values on one side and smaller ones on the other. This necessary condition is not sufficient.

Candidates include interior points where f′=0, interior points where the derivative does not exist and interval endpoints belonging to the domain. For a continuous function on a closed bounded interval [a,b], absolute maximum and minimum values exist. Compare the values of all these candidates. On open intervals, a limiting value may never be reached: f(x)=x on (0,1) approaches 1 but has no maximum. Distinguish a greatest existing value from an upper bound.

Justify an extremum and interpret curvature

If the derivative changes from positive to negative at a point, the function rises then falls: there is a local maximum. If it changes from negative to positive, there is a local minimum. If it keeps its sign, this criterion yields no extremum. For x³, the derivative 3x² is positive on both sides of zero, although it vanishes at that point.

The second derivative describes how the rate changes. Where f″>0, the derivative increases and the graph is concave upward; where f″<0, the reverse occurs. If f′(c)=0 and f″(c)>0, we have a local minimum; with f″(c)<0, a local maximum. When f″(c)=0, the test is inconclusive. Merely solving f″=0 does not establish an inflection point: a change in concavity must be confirmed and the point must belong to the graph.

An area model with a constraint

Imagine 48 m of fencing for a complete rectangle. If its sides are x and y, then 2x+2y=48, so y=24−x. The area is A=x(24−x), in m², with 0<x<24. The derivative A′=24−2x vanishes at x=12. It is positive before that value and negative afterwards. Therefore, the square of side 12 m reaches the greatest admissible area.

The verification can also be purely algebraic: A(x)=144−(x−12)². A square is always nonnegative, so A≤144, with equality only at x=12. This second solution confirms the conclusion and explains why balancing the sides improves the area. The derivative is a general tool, not an obligation when another structure solves the problem better. In geometric models, always seek this opportunity for independent verification.

Robust decisions and the model's limits

Mathematical optimisation depends on assumptions. If the fence includes a gate, if one side is already built or if only integer lengths are allowed, the constraint changes. A continuous optimum may serve as a reference, but a discrete choice requires comparing admissible neighbouring integers and any boundaries. Rounding without returning to the conditions may produce an impossible solution.

A complete answer presents the choice, optimal value, units and reason it is globally best. In applications, discuss sensitivity: near x=12, a change of 1 m in the first dimension, compensated in the other, reduces area by 1 m², since A(12+h)=144−h². Thus the model also informs the decision's tolerance. There is no need to calculate more decimal places than the data permit or to confuse numerical precision with physical accuracy.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Mistakes worth avoiding

All extrema satisfy f′=0.

Interval endpoints and nondifferentiable points are also candidates.

A zero derivative guarantees an extremum.

The sign may stay the same on both sides, as in x³.

The greatest value found is automatically admissible.

Check the domain and every model constraint.