The framework’s suggested skill for the introductory optimization topic is to identify common underlying structures in problems involving different contextual situations. That is an unusual thing to ask of a calculus topic, and it is the right thing to ask of this one. Fences, boxes, and distances are the same problem wearing different clothes.

The skeleton

Every one of these problems has four parts, in this order.

A quantity to optimize, named as a function. A constraint relating the variables, used to eliminate all but one of them. A domain for the surviving variable, forced by the situation rather than by the algebra. And only then the derivative.

The framework’s statement of the topic is deliberately plain: the derivative can be used to solve optimization problems, that is, finding a minimum or maximum value of a function on a given interval. The interval is given. Working out what it is, is the third step, and it is the step that gets skipped.

This is the same translation problem related rates poses, one unit later and in the other direction. There the prose hands you rates and asks for a rate; here it hands you a constraint and asks for an extreme value. Both stand or fall on getting a picture and a variable list onto paper before any calculus starts.

Three stories, one structure

On the left, the situation, redrawn as the free variable moves. On the right, the quantity being optimized plotted against that variable, with the feasible domain shaded and the candidates marked: filled dots for critical points, bars at the ends for endpoints. The panel carries the constraint, the objective, the domain, and every candidate with its value, which together are the whole write-up. The third problem has no endpoints at all, and its panel says so.

The fence and the box are the same problem. Each has a constraint that turns two variables into one, each produces a function on a closed interval, and in each the domain comes from geometry rather than algebra.

Notice where the two endpoints come from, because they come from different places. The left end is \(x = 0\) in both, because a width and a cut cannot be negative. The right end is whatever value collapses the other dimension: for the fence, \(y = 100 - 2x\) reaches zero at \(x = 50\); for the box, the base side \(12 - 2x\) reaches zero at \(x = 6\). Neither number appears anywhere in the algebra of the objective. Both have to be read off the picture.

For the fence, \(2x + y = 100\) gives \(A(x) = x(100-2x)\) on \([0,50]\), with \(A'(x) = 100 - 4x\) zero at \(x = 25\). The candidates are \(0\), \(25\), and \(50\), with areas \(0\), \(1250\), and \(0\). For the box, \(V(x) = x(12-2x)^2\) on \([0,6]\), and \(V'(x) = 12(x-2)(x-6)\) is zero at \(x = 2\) and \(x = 6\). The candidates are \(0\), \(2\), and \(6\), with volumes \(0\), \(128\), and \(0\) — and \(x = 6\) arrives twice, once as a critical point and once as an endpoint.

The Candidates Test, and when it does not apply

The framework states it in one line: absolute extrema of a function on a closed interval can only occur at critical points or at endpoints. So on a closed interval the whole method is to list those points, evaluate, and compare. No derivative test is needed, because comparing the values settles it directly — and the Extreme Value Theorem is what guarantees there is something to find, provided the objective is continuous on that interval.

Both parts of the definition of a critical point matter here. A critical point is where the derivative is zero or fails to exist, and objectives built from absolute values or roots can have the second kind without having the first.

The third problem is the one that breaks the habit. Minimising the distance from \((0,2)\) to the parabola \(y = x^2\) gives

\[D^2(x) = x^2 + (x^2-2)^2 = x^4 - 3x^2 + 4,\]

whose derivative \(2x(2x^2-3)\) vanishes at \(x = 0\) and \(x = \pm\sqrt{3/2}\). There is no closed interval and there are no endpoints, so there is nothing to compare against. Evaluating anyway: \(D^2 = 4\) at \(x = 0\) and \(D^2 = \tfrac74\) at the other two, so the minimum distance is \(\tfrac{\sqrt7}{2} \approx 1.3229\), attained twice.

And \(x = 0\) is a critical point that is a local maximum of the distance. A student who finds one critical point and stops has a defensible-looking answer of 2, which is wrong. Without endpoints to compare, the justification has to come from a derivative test, and here the second derivative \(6(2x^2-1)\) is \(-6\) at \(x = 0\) and \(+12\) at \(x = \pm\sqrt{3/2}\), which settles all three at once.

Answering the question that was asked

The suggested skill for the second optimization topic is to explain the meaning of mathematical solutions in context, which mostly means noticing what was asked for.

The fence question asks for the largest area, which is \(1250\) square metres, not \(x = 25\). The box question asks for the volume, \(128\) cubic inches, not the cut. And the distance question asks for a distance, so the answer is \(\tfrac{\sqrt7}{2}\), not \(\tfrac74\) — minimising \(D^2\) instead of \(D\) is a legitimate shortcut, because the square root is increasing and preserves the location of the minimum, but the shortcut has to be undone before the answer is written down.

The setup is worth practising apart from the calculus. Take a page of optimization problems and, for each one, write only three lines: the quantity being optimized, the constraint, and the domain with a reason for each end of it. Do not differentiate any of them. The differentiating is the part that already works, and separating the two makes it obvious how much of the difficulty lives in the first three lines.