Why Optimization Errors Often Survive the Calculus
In a constrained optimization problem, differentiating an objective is only one part of the work. A correct derivative cannot repair a missing physical restriction, an incorrect feasible interval, or a formula that measures the wrong quantity. Even after finding stationary points, you still have to compare all candidates that could give the requested optimum.
A reliable solution keeps three things connected: the feasible set, which describes the permitted choices; the objective function, which measures what is to be optimized; and the candidate comparison, which determines which permitted choice actually gives the largest or smallest value.
Start with Every Constraint, Not Just the Main Equation
A design problem often supplies an equation connecting dimensions, such as a volume or perimeter requirement. That equation is not necessarily the full set of constraints. Dimensions may also need to be nonnegative, a cut must fit inside a sheet, or a production quantity may be restricted to a specified range. These restrictions determine the feasible set and must be applied before interpreting the calculus.
Worked Example: Finding the Feasible Interval for an Open Box
A square sheet has side length \(20\) centimeters. Squares of side length \(x\) are cut from its corners, and the remaining sides are folded up to make an open box. Find the feasible interval for \(x\), then identify the volume-maximizing cut.
The box height is \(x\). Each base side is reduced by two cut lengths, so each has length \(20-2x\). Therefore the volume is
For a physical box, the cut length cannot be negative, and the remaining side length cannot be negative. Thus \(x\geq 0\) and \(20-2x\geq 0\), which gives \(x\leq 10\). The feasible interval, allowing degenerate boundary designs for comparison, is \(0\leq x\leq 10\). A nondegenerate box requires \(0<x<10\).
Differentiate the volume:
The stationary points satisfy \(20-2x=0\) or \(20-6x=0\), so \(x=10\) or \(x=10/3\). The endpoint and stationary-point values are
Thus the maximum occurs at \(x=10/3\) centimeters. The resulting base side is \(20-2(10/3)=40/3\) centimeters, and the height is \(10/3\) centimeters. Substituting these dimensions back into the volume formula gives \((10/3)(40/3)(40/3)=16000/27\) cubic centimeters, as required.
A concrete incorrect calculation is to try \(x=12\) because the formula \(V(x)=x(20-2x)^2\) is algebraically defined there. It gives \(V(12)=12(-4)^2=192\). But a base side of \(-4\) centimeters is impossible: \(x=12\) violates the constraint \(20-2x\geq 0\). An algebraically defined expression need not represent a feasible design.
Check That the Objective Measures the Requested Quantity
The constraint equation and the objective function play different roles. A volume constraint does not mean that volume is the quantity to minimize; a problem about material may require total surface area, including every surface made from material. Omitting a term can produce a plausible-looking function and a completely wrong design conclusion.
Worked Example: Including All Material in a Cylindrical Can
A closed cylinder must have volume \(1000\pi\) cubic centimeters. Find the radius that minimizes its total surface area. The cylinder has radius \(r>0\) and height \(h>0\). Its volume constraint gives
A closed can uses material for two circular ends and for its curved side. Its total surface area is \(S=2\pi r^2+2\pi rh\). Substituting the constraint gives the one-variable objective
Differentiating and setting the derivative equal to zero gives
So the stationary radius is \(r=\sqrt[3]{500}\) centimeters. At that radius the constraint gives \(h=1000/r^2=2r\), since \(r^3=500\). To verify that this is the global minimum, note that
and \(S(r)\) tends to infinity both as \(r\) approaches \(0\) from above and as \(r\) tends to infinity. The stationary point is therefore the unique global minimum. As a check, substituting \(h=2r\) into the full area formula gives \(2\pi r^2+2\pi r(2r)=6\pi r^2\).
A common mistake is to minimize only the curved-side area \(2\pi rh\). After substituting the volume constraint, that incomplete objective is \(2000\pi/r\), which decreases as \(r\) increases. It would suggest making the radius arbitrarily large. That conclusion ignores the two ends, whose combined area \(2\pi r^2\) grows with the radius. The corrected objective includes all the material specified in the problem.
Endpoints and Stationary Points Both Matter
For a continuous objective on a closed bounded interval, the extreme value theorem guarantees that a maximum and a minimum are attained. If the objective is also differentiable at every interior point where an extremum might occur, Fermat’s theorem says an interior extremum must have derivative zero. These facts give a practical candidate rule.
To justify the rule, continuity on the closed bounded interval ensures that \(f\) attains its global maximum and minimum. Let \(c\) be a point where either is attained. If \(c=a\) or \(c=b\), it is an endpoint candidate. Otherwise \(c\) is interior. For an interior maximum, sufficiently small positive and negative changes in the input cannot increase the function value, so the difference quotients from the two sides have opposite weak signs. If the derivative exists, the two-sided limit must therefore be zero. The same reasoning, with the signs reversed, applies to an interior minimum. Hence an interior extremum satisfies \(f'(c)=0\).
The rule does not say that every stationary point is an extremum, or that every local extremum is the global optimum. It also does not allow endpoints to be discarded. At a point where the objective is not differentiable, a possible extremum must be checked separately rather than ruled out by the derivative test.
Worked Example: Comparing a Stationary Point with the Endpoint
Suppose a design parameter \(q\) is restricted to \(0\leq q\leq 3\), and its score is \(P(q)=q^3-3q\). Find the value of \(q\) that maximizes the score.
Differentiate:
The stationary points of the formula are \(q=1\) and \(q=-1\), but only \(q=1\) lies in the feasible interval. The endpoint and feasible stationary-point values are
Therefore \(q=3\) gives the global maximum, with score \(18\). Substitution into the original objective confirms the value: \(P(3)=3^3-3(3)=27-9=18\).
The stationary point \(q=1\) is not the maximizing choice. Indeed, \(P''(q)=6q\), so \(P''(1)=6>0\), indicating a local minimum there. A wrong solution that keeps only the feasible stationary point would report a score of \(-2\), even though the allowed endpoint gives \(18\). The endpoint is part of the feasible set and must be compared.
A Reliable Diagnostic Checklist
Before accepting an optimization result, trace the argument from the original problem to the final choice. Each of the following checks catches a different class of error:
- Constraints: Have you included every stated requirement and every physical restriction, such as nonnegative lengths?
- Feasible set: Does each candidate satisfy all constraints? Are the interval endpoints included, excluded, or physically degenerate?
- Objective: Does the function measure exactly the requested quantity, including all relevant costs, areas, or components?
- Candidate list: Have you considered endpoints, interior stationary points, and any feasible points where the objective is not differentiable?
- Comparison: Have you evaluated the original objective at every candidate and selected the largest or smallest value as requested?
- Verification: Do the final dimensions or decision values satisfy the original constraints when substituted back?
The interval theorem above assumes a closed bounded interval. If a feasible interval is open, an endpoint may not be an allowed choice; values approached near that endpoint can influence the behavior without being attained. In that situation, do not treat an excluded endpoint as a feasible candidate or claim a maximum exists merely because nearby values are large. First determine what choices are actually permitted, then establish whether the required optimum is attained.
Check Your Understanding
For each question, explain which part of the optimization argument needs checking.
- A cut-box formula has a stationary point at a cut length larger than half the sheet’s side. What must be checked before interpreting that point?
- A model minimizes the area of a closed container but includes only the curved side. Which part of the model is incomplete?
- On a closed interval, why is it not enough to evaluate only the interior points where the derivative is zero?
- What does a stationary point tell you, and what does it not tell you, about a constrained maximum?
- Why should a proposed design be substituted into the original constraint and objective before the answer is accepted?
