Compare all candidates before choosing a design

Try

A builder has 60 metres of fencing for three sides of a rectangle, with a wall supplying the fourth side. Let w be each perpendicular side. Increasing w makes the rectangle deeper but leaves less fence for its long side. Predict why maximizing one dimension alone cannot maximize area.

Predict

Predict whether solving A′(w)=0 is enough to prove a global maximum. What if the allowed width interval excluded the candidate? What if the best value occurred at an endpoint or a corner where the derivative did not exist? The feasible domain is part of the optimization problem.

Experiment

The remaining side is 60−2w, so A(w)=w(60−2w)=60w−2w² square metres. Allow the closed interval 0≤w≤30 to include degenerate zero-area designs while comparing candidates. Positive-area rectangles use the interior.

Candidate width	Remaining side	Area
0	60	0
15	30	450
30	0	0

The derivative is 60−4w, zero at 15. It is positive before that input and negative afterward, and the second derivative is −4. Sketch the area parabola only over the feasible interval. The sign and endpoint comparisons jointly support the maximizing design.

Understand

The extreme value theorem says a continuous real function on a closed bounded interval attains both a maximum and a minimum. It guarantees existence, not their locations. To find them, compare endpoints and all interior critical candidates: points where the derivative is zero or fails to exist while the function itself exists. Some functions have infinitely many candidates, requiring a more analytical comparison rather than a finite checklist.

For f(x)=x³−3x on [−2,3], the stationary inputs are −1 and 1. Evaluate f at −2, −1, 1 and 3: the values are −2, 2, −2 and 18. The local maximum at −1 is not the interval’s global maximum; the right endpoint is larger. This is why a local classification cannot replace the global comparison.

For |x−2| on [0,5], the minimum occurs at the corner x=2. Looking only for derivative zeros misses it entirely. For f(x)=x on the open interval (0,1), there is no attained maximum or minimum: the supremum 1 and infimum 0 are approached but excluded. Bounded values and continuity without a closed interval do not supply the theorem’s conclusion.

Applied optimization begins before differentiation. Define variables and units, express the objective, translate constraints and specify the feasible domain. For an open box cut from a 12 by 12 sheet, removing corner squares of side x gives V=x(12−2x)². Differentiation gives (12−2x)(12−6x), so the interior candidate is x=2. Endpoint volumes at 0 and 6 are zero and V(2)=128 cubic units. These comparisons establish the maximum among the stated idealized designs.

Real design choices may require integer dimensions, thickness, safety margins or multiple competing objectives. The continuous model gives evidence under its stated assumptions, not an automatic practical recommendation. If quantities must be discrete, compare feasible neighboring values and any boundaries instead of rounding without checking.

Apply

For each practice problem, write the objective and feasible set before calculating derivatives. Distinguish “maximizing input” from “maximum output value.” In a theorem response, state attainment, not merely boundedness. Return to the tangent investigation to examine nondifferentiable candidates and why they cannot be ignored.

When a local extremum is the global one

Let f be continuous on an interval I, and suppose its only interior critical point is c: at every other interior point the derivative exists and is nonzero. If c is a local maximum, then it is a global maximum on I. This statement does not replace the hypotheses with “the derivative equation has one root”: undefined derivatives are also critical candidates.

Here is a proof without assuming a bounded interval or endpoint attainment. If some b>c in I had f(b)>f(c), choose a strictly between c and b close enough to c that local maximality gives f(a)≤f(c). Equality is impossible: Rolle’s theorem on [c,a] would produce another critical point. Thus f(a)<f(c). By EVT, f attains a minimum on [c,b]; its value is at most f(a), strictly below both endpoint values. The minimum is therefore attained at an interior point different from c, and Fermat’s theorem makes that a second critical point, a contradiction. The same argument works if b<c. Replace f by −f for a local minimum. Notice where continuity, the local property and absence of other critical points were each needed.

Transfer. For f(x)=x+4/x on (0,∞), show that its unique critical point is a global minimum, although the domain is open. Then explain why the analogous inference from “one critical point” alone fails for g(x)=x³ on the real line.

Review the hypotheses, not just a derivative root

Here f′=1−4/x² vanishes only at x=2, is negative before 2 and positive after, giving a local minimum. Continuity on (0,∞) and differentiability elsewhere meet the theorem; the global minimum is f(2)=4. No endpoint comparison at zero is legitimate because zero is outside the domain. For g, the derivative 3x² vanishes only at zero, but zero is not a local extremum. The missing local-extremum hypothesis blocks the conclusion. Credit requires the sign or another valid local-extremum argument as well as uniqueness of the critical point.

Continue

The final lesson addresses limits where substitution is indeterminate. Differentiation can help, but replacing a ratio by a derivative ratio requires a theorem with specific hypotheses, not a general algebraic identity.