Constrained optimization

Derive and check a constrained optimum rather than treating a first-order condition as sufficient.

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1. What you will learn

You will derive and check a constrained optimum rather than treating a first-order condition as sufficient, showing the calculation and stating the assumptions that make the conclusion valid.

2. Starting point

Use the supplied definitions and units. Separate an accounting identity, a behavioral assumption, and a normative criterion before drawing conclusions.

3. Terms to use precisely

4. An optimization problem has three distinct ingredients

An economic optimization problem specifies an objective, a feasible set, and a choice variable. The objective tells us how the model ranks outcomes. The feasible set identifies choices consistent with resource, technological, institutional, or other constraints. The choice variable records what the decision maker can adjust. Leaving one of these ingredients implicit can turn a valid calculation into an answer to a different question.

Consider a bounded production-intensity model. The decision maker chooses q between zero and a capacity K. Its objective is F(q)=a q-b q squared, with b positive. The linear term represents the model's initial gain from increasing q, while the quadratic term makes the additional gain decline. The expression is a stipulated net objective; it is not an empirical law that every producer or household has quadratic preferences or costs.

The constraint zero at most q at most K matters independently of the objective. An unconstrained maximizer might require more capacity than is available or a negative quantity that is not meaningful for this decision. Solving the objective's derivative equation without checking feasibility would therefore be incomplete. A stationary point is a candidate from the mathematical expression, not permission to choose an infeasible quantity.

We use continuous q in this lesson so calculus describes local changes. If quantities must be whole items, the optimal feasible integer must be selected separately, often by comparing nearby integers and relevant boundaries. A continuous solution of 2.5 does not authorize ordering half an indivisible machine. State the domain before treating a first-order condition as a complete economic recommendation.

Term	What it means
Feasible set	Choices satisfying the model's stated constraints.
Stationary point	A differentiable interior candidate where the objective derivative is zero.
Strict concavity	Curvature that makes a stationary point a unique maximum over a convex feasible set when the relevant conditions hold.
Corner solution	An optimum at a boundary of the feasible set.
Shadow value	The local change in the optimized objective from relaxing a specified constraint.
Complementary slackness	The condition that a nonzero inequality multiplier is associated with a binding constraint.

Another way: Derive the stationary point and check curvature

For F(q)=a q-b q squared, the derivative is a-2bq. It measures the model's local change in the objective from a small increase in q. Setting that derivative equal to zero gives the unconstrained stationary point q=a/(2b). The division is legitimate because b is positive. This step locates where the marginal gain changes sign in the quadratic model.

The second derivative is negative 2b, which is strictly negative. The objective is therefore strictly concave over the real line: its slope declines, and the stationary point is the unique unconstrained maximum. This curvature check supplies the justification that a zero derivative alone lacks. For a different objective such as positive q squared, the zero derivative at zero would identify a minimum rather than a maximum.

A negative second derivative at one point is a local condition; global conclusions require the relevant behavior over the feasible set or another sufficient argument. Our quadratic has the required concavity everywhere, and the feasible interval is convex. Those assumptions make the constrained maximum unique except for degenerate cases excluded by b positive. In a nonconcave model, multiple stationary points and boundaries may need explicit comparison.

Always distinguish necessity from sufficiency. For a differentiable interior optimum, a zero derivative is necessary under the usual conditions. At a constrained boundary, the derivative need not be zero. And without suitable curvature or global comparison, a stationary point need not be the best feasible choice. Economic explanations should say which of these claims is being used instead of treating 'set derivative to zero' as a universal proof.

Another way: Check both boundaries as well as the interior

If a/(2b) lies between zero and K, it is feasible and gives the maximum in this concave problem. If it lies above K, the objective is still increasing over the feasible interval up to K, so the capacity boundary is optimal. If it lies below zero, the objective is decreasing over the nonnegative feasible interval, so zero is optimal. We can summarize these cases by clipping the unconstrained candidate to the interval.

The summary rule is useful because it has been justified for this specific strictly concave quadratic and interval constraint. It is not a general instruction to clip any stationary point in every optimization problem. In a nonconcave problem, an interior local maximum or the opposite boundary could have a larger value. Derive a rule from the assumptions before applying it to a new functional form.

At an upper-bound optimum, the derivative can remain positive. Suppose a is twelve, b one, and K four. The unconstrained candidate is six, but only quantities up to four are feasible. At q four the derivative is four, indicating that additional capacity would be valuable locally. The positive derivative does not show that the chosen feasible point is wrong; it shows the constraint prevents the preferred unconstrained expansion.

At a lower-bound optimum, the derivative can be negative. If a is negative and b positive, any positive q reduces the objective relative to zero. Choosing zero is then a substantive corner solution, not a failure to solve the equation. Economic constraints such as nonnegative consumption or output often make boundary analysis essential rather than an exceptional afterthought.

Another way: A shadow value describes a local relaxation

The upper-capacity shadow value measures the local improvement in the optimized objective when K increases slightly, holding the other model parameters fixed. In this quadratic interval model, it is a-2bK when the upper constraint binds below the unconstrained optimum, and zero when additional capacity does not change the chosen quantity. The units are objective units per additional capacity unit.

For a twelve, b one, and K four, the shadow value is four. A small extra amount of capacity would improve the maximum objective at approximately four times that small amount. This is a local rate, not an exact prediction that every large capacity expansion yields four per unit. As capacity rises toward six, the marginal value declines, then becomes zero once the unconstrained optimum is feasible.

The distinction can be expressed with a Lagrangian. Using F(q)+lambda(K-q)+mu q, nonnegative lambda and mu correspond to the upper and lower inequality constraints. Stationarity gives a-2bq-lambda+mu=0. Complementary slackness requires lambda(K-q)=0 and mu q=0. A positive multiplier can therefore attach only to a binding constraint, although a binding constraint can have a zero multiplier at a transition.

These conditions are interpreted within the model. A shadow value is not automatically a market price, a willingness to pay in money, or a policy recommendation. Its meaning depends on the objective's units, the local nature of the change, and the held-fixed parameters. If expanding capacity itself changes costs or technology, the simple relaxation calculation no longer captures the whole proposal.

Another way: Check the result and distinguish prediction from evaluation

After finding q, substitute it into both the constraint and the objective. The chosen quantity must lie in the allowed interval, and the reported objective value must use that feasible quantity rather than the unconstrained candidate. This second substitution catches a common mistake: correctly identifying that capacity binds but reporting the higher payoff that only the infeasible unconstrained choice would achieve.

Then verify the marginal interpretation. An interior solution should have derivative zero. An upper corner should have a nonnegative derivative in this concave model, while a lower corner should have a nonpositive derivative. The sign conditions explain why no feasible small movement can improve the result. They connect the formula to a reasoned account of the constrained choice.

For a finite change in capacity, solve the new optimization problem rather than multiplying the old shadow value across the entire change. For an integer-choice version, compare the feasible candidate quantities directly. For an uncertain parameter, distinguish the assumed value from evidence that supports it. Each modification changes a part of the original problem and therefore requires checking which derivation still applies.

Finally, optimization is conditional on an objective. A profit maximum, utility maximum, or surplus maximum answers a different economic question, even if the mathematical method is similar. The calculation does not establish that the chosen objective fully represents wellbeing or that observed people always solve the model. It supplies a transparent implication: if this objective, feasible set, and parameterization govern the problem, this choice is best within them.

Another way: Comparative statics hold other parameters fixed

In an interior solution, increasing a raises the chosen quantity because a/(2b) rises, while increasing b lowers it. At a binding upper capacity, a modest increase in a may leave the chosen quantity unchanged even though it raises the shadow value of capacity. The response therefore depends on which constraint is active. Comparative statics means changing a specified parameter while holding the other components fixed; it is not a claim that real-world variables always move independently. When several parameters change together, solve the revised problem and identify the separate mechanisms before attributing the observed response to one cause.

5. A fictional workshop's capacity decision

A fictional workshop evaluates intensity q using the stipulated net objective F(q)=12q-q squared. Quantities are continuous for the exercise, q cannot be negative, and current capacity limits q to four. The model is deliberately compact: its objective already incorporates the costs named by the case, and no additional market or welfare claim is inferred from its coefficients.

The derivative is twelve minus two q, giving an unconstrained candidate of six. The second derivative is negative two, so the objective is strictly concave. Six is infeasible under capacity four. The optimum is therefore q four, with objective forty-eight minus sixteen, or thirty-two. Reporting thirty-six would mistakenly substitute the infeasible q six into the payoff calculation.

At q four, the derivative and upper-capacity shadow value are four objective units per unit of capacity. If capacity rises to five, the new optimum is five and the objective becomes sixty minus twenty-five, or thirty-five. The actual gain is three, not four: the old shadow value was a local rate, and marginal value declines across the full one-unit expansion.

If capacity instead rises to seven, the workshop chooses the unconstrained optimum six and obtains thirty-six. The extra seventh unit of capacity has zero local value under the unchanged objective. These comparisons show how to check feasibility, curvature and payoffs together. They do not recommend a real investment, because the exercise has not supplied the cost of acquiring capacity or evidence that the assumed objective describes an actual workshop.

6. Check the tempting inference

A zero derivative is not sufficient without curvature or a global comparison, and a constrained optimum need not have zero derivative. Apply both bounds before computing the payoff. Shadow values are local and inherit the objective's units; they do not automatically equal market prices or justify an investment.

7. A feasible interior maximum

8. Capacity binds before the stationary point

9. The lower boundary and a multiplier check

10. Finish a capacity-limited optimum

11. Guided practice

Maximize F(q)=16q-q^2 over continuous 0<=q<=10. Give optimal q, maximum objective and local upper-capacity shadow value.

12. Guided practice

Maximize F(q)=8q-q^2 over 0<=q<=3. Complete the candidate and feasible-choice calculation.

13. Guided practice

Maximize F(q)=14q-q^2 over 0<=q<=4. Give optimal q, attainable maximum and local upper-capacity shadow value.

14. Practice

Maximize F(q)=-6q-q^2 over continuous 0<=q<=8. Produce the optimal nonnegative q only.

15. Practice

Maximize F(q)=24q-2q^2 over 0<=q<=9. Give optimal q, maximum objective and upper-capacity shadow value.

16. Somewhere new

A fictional workshop's stipulated net objective is 18q-q^2 with continuous intensity and capacity 0<=q<=5. Reconstruct its optimal q, attainable objective and local value of relaxing the capacity bound, without inferring a real investment recommendation.

17. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

18. Test question

A fresh bounded model maximizes F(q)=20q-2q^2 on continuous 0<=q<=3. Give optimal q, attainable maximum and upper-capacity shadow value.

19. What you can do now

Reconstruct a fresh case without the worked solution. Explain which assumption would change its conclusion and which result is only an accounting or model condition.

	Your result
Optimal q
Maximum objective
Upper-capacity shadow value