Checking Constrained Optima With KKT Conditions

Why KKT Conditions Matter for Pricing

A price that maximizes an unconstrained objective may not be available to a decision-maker. A permitted price range, such as a floor and ceiling, can exclude that price and move the optimum to one of the boundaries. The Karush-Kuhn-Tucker (KKT) conditions provide a systematic way to check whether a proposed constrained price is optimal and to identify which bounds are active.

This tutorial considers one differentiable objective $f(p)$, where $p$ is the price, subject only to a lower bound $L$ and an upper bound $U$, with $L<U$. The objective may represent a hypothetical insurer’s pricing objective; its units and interpretation depend on how $f$ is defined. The method applies whether the objective is profit, revenue, or another differentiable quantity, but the objective must be specified before optimizing.

Definition: For a price constrained by $L\le p\le U$, the lower bound is binding at $p$ if $p=L$, and the upper bound is binding if $p=U$. A bound is slack if its inequality is strict. A Lagrange multiplier associated with a bound measures the marginal change in the optimized objective when that bound is relaxed, under the regularity and differentiability conditions for the sensitivity interpretation.

KKT Conditions for Price Bounds

Write the two constraints in the standard form $g_j(p)\le 0$:

g_L(p)=L-p\le 0,\qquad g_U(p)=p-U\le 0.

For a maximization problem with constraints in this form, use nonnegative multipliers $\lambda_L$ and $\lambda_U$, and form the Lagrangian

\mathcal{L}(p,\lambda_L,\lambda_U) =f(p)-\lambda_L(L-p)-\lambda_U(p-U) =f(p)+\lambda_L(p-L)-\lambda_U(p-U).

The KKT conditions are stationarity, primal feasibility, dual feasibility, and complementary slackness. For a differentiable objective and ordinary interval bounds, they are:

\begin{aligned} &f'(p)+\lambda_L-\lambda_U=0 &&\text{(stationarity)},\\ &L\le p\le U &&\text{(primal feasibility)},\\ &\lambda_L\ge 0,\quad \lambda_U\ge 0 &&\text{(dual feasibility)},\\ &\lambda_L(p-L)=0,\quad \lambda_U(U-p)=0 &&\text{(complementary slackness)}. \end{aligned}

Complementary slackness connects each multiplier to its constraint. If $p>L$, then $\lambda_L=0$; if $p<U$, then $\lambda_U=0$. A constraint can be binding while its multiplier is zero in a degenerate case, so a binding constraint does not always imply a strictly positive multiplier.

Why the Boundary Signs Have This Form

At an interior maximum, small changes in either direction are feasible, so the derivative must be zero. At the lower bound, only small increases in price are feasible. If $f'(L)>0$, increasing price slightly increases the objective, so the lower bound cannot be optimal. Thus a lower-bound maximum requires $f'(L)\le 0$. At the upper bound, only small decreases are feasible; if $f'(U)<0$, decreasing price slightly raises the objective. Thus an upper-bound maximum requires $f'(U)\ge 0$.

These signs also follow directly from KKT stationarity. At the lower bound, complementary slackness gives $\lambda_U=0$, so $f'(L)+\lambda_L=0$, or $\lambda_L=-f'(L)\ge0$. At the upper bound, $\lambda_L=0$, so $f'(U)-\lambda_U=0$, or $\lambda_U=f'(U)\ge0$. This establishes the boundary tests and explains why the multiplier signs depend on the chosen constraint convention.

Theorem: If $f$ is differentiable and concave on the feasible interval $[L,U]$, any feasible price satisfying the KKT conditions is a global maximizer. If $f$ is strictly concave, that maximizer is unique. For a nonconcave objective, KKT conditions generally identify candidates, not a guaranteed global maximum; compare feasible candidates and endpoints.

For price bounds with $L<U$, the conditions can be read as a concise decision rule: an interior candidate has $f'(p)=0$; a lower-bound candidate requires $f'(L)\le0$; and an upper-bound candidate requires $f'(U)\ge0$. These derivative checks are especially useful when the unconstrained optimum lies outside the permitted interval.

Worked Example: A Ceiling That Binds

A hypothetical pricing objective, measured in thousands of dollars, is $f(p)=(p-40)(100-p)$, where $p$ is the annual premium in dollars. The permissible interval is $60\le p\le65$. Find the constrained optimum and its KKT multipliers.

Method. Expand and differentiate the objective:

f(p)=-p^2+140p-4000,\qquad f'(p)=140-2p,\qquad f''(p)=-2.

Since $f''(p)=-2<0$, the objective is strictly concave. Its unconstrained stationary price solves $140-2p=0$, giving $p=70$, which exceeds the ceiling. On the feasible interval, the derivative ranges from $f'(60)=20$ to $f'(65)=10$, so the objective is increasing throughout the interval. The constrained maximum is therefore at $p=65$.

At $p=65$, the lower bound is slack, so complementary slackness gives $\lambda_L=0$. Stationarity gives

10+0-\lambda_U=0,\qquad \lambda_U=10.

The upper-bound multiplier is nonnegative, and the price is feasible, so all KKT conditions hold. The objective value is $f(65)=(65-40)(100-65)=25(35)=875$, or $875,000 under the stated units. As a check, the other endpoint gives $f(60)=20(40)=800$; the objective increases from 800 to 875 over the allowed interval. The common mistake is to use the unconstrained price $70$ even though it violates the price ceiling.

Worked Example: A Floor That Binds

A second hypothetical objective is $f(p)=(p-30)(90-p)$, again in thousands of dollars. The allowed price range is $65\le p\le80$. Determine the maximizing price, its multiplier, and the objective value.

Expand and differentiate:

f(p)=-p^2+120p-2700,\qquad f'(p)=120-2p,\qquad f''(p)=-2.

The unconstrained stationary point is $120-2p=0$, or $p=60$, below the floor. On the feasible range, $f'(65)=-10$, and the derivative decreases further to $f'(80)=-40$. Thus the objective is decreasing throughout the allowed interval, making $p=65$ the constrained maximum.

At the floor, the upper bound is slack, so $\lambda_U=0$. Stationarity gives

-10+\lambda_L-0=0,\qquad \lambda_L=10.

The lower-bound multiplier is nonnegative. The value at the optimum is $f(65)=(65-30)(90-65)=35(25)=875$, or $875,000. As a check, the value at $p=80$ is $50(10)=500$, which is lower. The common mistake is to reject a boundary because the derivative is not zero: at a constrained floor, the required condition is $f'(L)\le0$, not $f'(L)=0$.

Worked Example: An Interior Optimum With Slack Bounds

Suppose a hypothetical pricing objective, in thousands of dollars, is $f(p)=1200-(p-72)^2$, with permitted prices $60\le p\le85$. Verify the optimum using KKT conditions and check the objective value.

Differentiate to obtain $f'(p)=-2(p-72)$ and $f''(p)=-2<0$. The stationary point is $p=72$, which is strictly between 60 and 85. Both bounds are slack, so complementary slackness requires $\lambda_L=\lambda_U=0$. Stationarity is satisfied because

$$ f'(72)+0-0=-2(72-72)=0. $$

The price is feasible, both multipliers are nonnegative, and all KKT conditions hold. Strict concavity establishes that $p=72$ is the unique global maximum. Its objective value is $f(72)=1200-0^2=1200$, or $1,200,000. The endpoint values are $f(60)=1200-144=1056$ and $f(85)=1200-169=1031$, both lower. The common mistake is to assign a positive multiplier to a slack bound; complementary slackness makes both multipliers zero here.

A Reliable Checking Sequence

A candidate should not be accepted merely because its price looks plausible or its derivative has the expected sign. Check feasibility, the applicable stationarity equation, and each multiplier and complementary-slackness condition. Concavity then determines whether the candidate is a global maximum.

Write the feasible interval.
Record $L\le p\le U$ and confirm that the proposed price lies within it.

Differentiate the objective.
Compute $f'(p)$, and inspect curvature when possible. Concavity supports a global-optimality conclusion.

Classify each bound.
At an interior price both bounds are slack; at $L$ the lower bound binds; at $U$ the upper bound binds.

Apply KKT and verify.
Use stationarity and complementary slackness to calculate multipliers, check their signs, and evaluate the objective if needed.

Common KKT Errors in Pricing

Using the wrong multiplier sign. With $g(p)\le0$ for a maximization problem and $\mathcal{L}=f-\lambda g$, multipliers must be nonnegative. Changing the constraint or Lagrangian convention changes the written stationarity equation, so state the convention first.
Requiring zero derivative at a boundary. A constrained maximum at the lower bound can have a negative derivative; one at the upper bound can have a positive derivative. Those signs point outward from feasible improvements.
Confusing binding with positive multiplier. Complementary slackness says a slack constraint has zero multiplier. A binding constraint may still have a zero multiplier in a degenerate case.
Calling a KKT point a global optimum without checking curvature. Concavity makes KKT sufficient for global optimality. Without concavity, compare the feasible candidates and endpoints rather than assuming sufficiency.
Ignoring the constraint when using an unconstrained solution. An unconstrained stationary price outside $[L,U]$ is not a feasible answer. Check which boundary is optimal using the derivative or the objective values.

KKT conditions organize the logic of constrained pricing: feasibility says the price can be used, stationarity balances the objective’s slope against active bounds, and complementary slackness ensures that slack bounds do not exert a force on the solution. The multiplier can also quantify the local value of relaxing a binding bound, but it is not automatically the total change from a large policy change.

Check Your Understanding

For each problem, identify the maximizing price and, where relevant, the nonzero KKT multiplier. Assume each objective is measured in the units stated and that the price is restricted to the given interval.

Maximize $f(p)=400-(p-50)^2$ over $45\le p\le55$. State whether either bound binds. (Answer: $p=50$; both bounds are slack and $\lambda_L=\lambda_U=0$.)
Maximize $f(p)=900-(p-62)^2$ over $50\le p\le58$. Find the active bound and its multiplier. (Answer: $p=58$; the upper bound binds, $\lambda_U=8$, and $\lambda_L=0$.)
Maximize $f(p)=600-(p-40)^2$ over $45\le p\le70$. Find the optimum and the lower-bound multiplier. (Answer: $p=45$; $\lambda_L=10$ and $\lambda_U=0$.)
Maximize $f(p)=-p^2+100p$ over $30\le p\le60$. Find the price, objective value, and multiplier status. (Answer: $p=50$, $f(50)=2500$; both bounds are slack and both multipliers are zero.)
Maximize $f(p)=-2p^2+200p$ over $52\le p\le70$. Verify the constrained optimum using the derivative sign and give its multiplier. (Answer: $p=52$; $f'(52)=-8$, so the lower bound binds with $\lambda_L=8$, and $\lambda_U=0$.)

What You'll Learn