Calculus I · Unit 2B · lesson
Closed-Interval Optimization
Learning objectives
Use the closed interval method in optimization models and explain endpoint meaning.
Compare Every Feasible Candidate
Before the formulas
The first job in Closed-Interval Optimization is modeling, not differentiation. Name the quantity to optimize, write the constraint, use the constraint to reduce the objective to one variable, and determine the physically feasible domain. Only then should you take a derivative.
A critical point is a candidate, not the answer. Verify that it lies in the domain and compare it with endpoints or use an appropriate sign or concavity argument. Finish by answering the original question with units and dimensions, not merely reporting the value of the variable used in the derivative.
Optimization on a bounded domain is a comparison problem
When the feasible domain is a closed interval, use the closed-interval method: evaluate the objective at endpoints and interior critical numbers. The derivative finds where improvement stops locally; endpoint evaluation checks whether the best feasible choice occurs at a boundary.
Always report the requested quantity, not merely the variable used in the derivative. If the problem asks for dimensions, give all dimensions; if it asks for maximum profit, compute the profit at the optimizing production level.
Many practical optimization problems have hard limits: a dose cannot be negative, a machine has a maximum setting, and a material sheet has fixed dimensions. On a closed feasible interval, compare endpoints and interior critical points to find the true absolute optimum.
A local optimum may be irrelevant if an endpoint performs better. The original objective function, not the derivative value, determines the winner.
Many physical optimization functions are continuous on a feasible closed interval. The Extreme Value Theorem guarantees extrema, and the closed interval method identifies them.
Open-top box from a sheet
Squares of side are cut from each corner of a -by- cm sheet and the sides are folded up. Find the cut size that maximizes volume.
Worked solution
Write a real attempt before opening the supplied answer.
An algebraic critical number outside the feasible domain is not a physical candidate. A negative length or a cut larger than half the shorter side is not an exotic box; it is a modeling error.
A constrained medication response
Suppose a response score is
for doses . The derivative gives the interior critical point . Evaluate
The absolute maximum on the permitted interval occurs at . The endpoint comparison matters because safety constraints can exclude unconstrained behavior beyond the interval.
After the explanation
Use the section idea
Separate the objective from the constraint, reduce to one feasible variable, and interpret the winning candidate in the original design.
Optimization is a modeling problem first: the derivative only compares candidates after the geometry, units, and feasible domain are correct.
Write variables and units, identify the objective, use the constraint to eliminate a variable, then test critical and boundary candidates.
Optimizing the constraint, ignoring the feasible domain, or keeping an algebraic critical point that cannot exist in the real design.
Have you compared every feasible candidate and explained why the result is physically and economically reasonable?
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