Topic 5.11 Solving Optimization Problems: solve a complete optimization problem and justify the absolute extremum.

What this topic is asking

The College Board (Topic 5.11) is the solving half of optimization: differentiate the one-variable objective from Topic 5.10, find the critical point, and crucially justify that it is the absolute maximum or minimum the question asks for. The justification is where most marks are won or lost.

The solving procedure

A worked optimization

Justifying "absolute" on an open domain

Many optimization domains are open (like $x > 0$), so the Extreme Value Theorem does not directly apply and you cannot simply use the candidates test with endpoints. The clean justification is then the Second Derivative Test giving constant concavity, or a First Derivative Test sign change that holds throughout: if $Q'$ changes from negative to positive at the only critical point, that point is the absolute minimum on the interval. State this explicitly. A bare "$S'(x) = 0$ at $x = 4$" earns the critical-point mark but not the justification mark; the exam separately rewards the argument that the critical point is the global extremum.

Reading exactly what is asked

Optimization questions vary in what they want: the optimal dimensions, the optimal value of the objective, or both. A question asking for the "minimum surface area" wants the value $S$, not the side $x$; one asking "what dimensions minimize the cost" wants the $x$ (and any other dimensions). Compute the critical $x$ either way, then read the question to decide whether to substitute back to get the objective value. Carrying units through to the final answer and writing a short sentence in context complete a full-credit free response. The most common avoidable loss is finding $x$ correctly but never justifying that it is the absolute extremum.

Choosing the cleanest justification

The three justification routes are not equally convenient on every problem, and choosing well saves time. When the objective's domain is a closed interval, the candidates test is fastest: just compare the objective at the critical point and the two endpoints. When the domain is open but the second derivative is easy to compute, the Second Derivative Test gives the answer in one evaluation, and if $S''$ has a constant sign on the whole domain you have established a global, not merely local, extremum. When $S''$ is messy, a First Derivative Test sign change is the safer route. Picking the lightest valid justification for the situation, rather than mechanically computing $S''$ every time, keeps the algebra manageable on the no-calculator section.

Common problem types to recognize

A handful of templates recur: maximizing area or volume for a fixed amount of material, minimizing surface area or cost for a fixed volume, minimizing the distance from a point to a curve, and maximizing a revenue or profit function. Recognizing the type tells you what the objective and constraint usually are, a fixed perimeter or volume as the constraint, the area, volume, or cost as the objective, so the setup goes faster. The calculus that follows is always the same: differentiate, solve $S'(x) = 0$, and justify the absolute extremum. Familiarity with the templates frees attention for the parts that vary, the specific geometry and the domain, where the marks are actually decided.