Utility-Maximizing Dose from Quadratic Response Models

Why a Quadratic Model Gives a Dose-Finding Rule

When benefit initially rises with dose but eventually levels off or declines, while harm may increase more rapidly at higher doses, the best dose need not be the largest one available. A quadratic response model represents this changing pattern with a relatively simple function. If benefit and harm are combined into a utility function, calculus can identify a candidate optimum directly. That candidate is only a recommendation after it has been checked against the feasible dose range.

The goal here is to optimize utility at fixed model parameters, as in the earlier material on choosing a dose by expected utility. The calculation does not account for uncertainty in estimated parameters. It answers a conditional question: given these response curves, utility weights, and feasible doses, which dose has the greatest modeled utility?

Definition: Let $d$ be dose in a feasible interval $[d_{\min},d_{\max}]$. Suppose the modeled benefit $B(d)$ and harm burden $R(d)$ are quadratic functions: $$ B(d)=b_0+b_1d+b_2d^2,\qquad R(d)=r_0+r_1d+r_2d^2. $$ For nonnegative utility weights $w_B$ and $w_R$, define $$ U(d)=w_BB(d)-w_RR(d). $$ A utility-maximizing dose is any feasible $d$ at which $U(d)$ is at least as large as at every other feasible dose.

The minus sign assumes that a larger value of $R(d)$ represents more harm. The weights translate the benefit and harm scales into a common utility scale; they must reflect the chosen preference structure. An additive constant in utility does not change which dose maximizes it, but changing a weight generally can change the maximizing dose. The response functions may represent expected outcomes, but the algebra below requires only that the specified utility is the function being optimized.

Combine the Response Models Before Optimizing

Substituting the two quadratic models into utility shows that utility is itself quadratic:

U(d)=q_0+q_1d+q_2d^2, \quad q_0=w_Bb_0-w_Rr_0,\quad q_1=w_Bb_1-w_Rr_1,\quad q_2=w_Bb_2-w_Rr_2.

The coefficient $q_1$ is the utility slope at dose zero, if zero is on the model's dose scale. The coefficient $q_2$ controls curvature: negative $q_2$ means utility bends downward, positive $q_2$ means it bends upward. These are combined coefficients. For example, even if benefit increases with diminishing returns, the net utility may curve upward if the weighted harm response bends upward strongly enough.

Units should be checked before interpreting coefficients. If dose is measured in milligrams and utility in utility units, $q_1$ has units of utility units per milligram, while $q_2$ has units of utility units per milligram squared. Coefficients from a model using a different dose unit cannot be substituted without converting the dose scale consistently.

Result: If $q_2<0$, the unconstrained quadratic utility has a unique maximum at $$ d^*=-\frac{q_1}{2q_2}. $$ For a closed feasible interval, this is the constrained maximum if it lies within the interval. If it lies outside, the constrained maximum is at the nearer endpoint. If $q_2>0$, the stationary point is a minimum, and the maximum on a nondegenerate closed interval is at one of its endpoints. If $q_2=0$, utility is linear or constant, so its maximum is at an endpoint or at every dose, respectively.

Why the Stationary Dose Is Not Always the Answer

Differentiate the utility function with respect to dose:

\frac{dU(d)}{dd}=q_1+2q_2d.

Setting this derivative to zero gives $d^*=-q_1/(2q_2)$, provided $q_2\ne 0$. The second derivative is $2q_2$. Thus, if $q_2<0$, the curve is strictly concave and the stationary point is a maximum; if $q_2>0$, it is strictly convex and the stationary point is a minimum. The derivative alone does not establish that the point is feasible, nor does a stationary point exist when utility is linear with nonzero slope.

For a concave quadratic, completing the square makes the boundary rule explicit:

U(d)=q_2\left(d+\frac{q_1}{2q_2}\right)^2 +q_0-\frac{q_1^2}{4q_2} = q_2(d-d^*)^2+U(d^*).

When $q_2<0$, the squared term is multiplied by a negative number. Utility is therefore greatest at the feasible dose closest to $d^*$. If $d^*$ falls below $d_{\min}$, utility decreases throughout the feasible interval and the lower boundary is best. If it exceeds $d_{\max}$, utility increases throughout the interval and the upper boundary is best. This is why checking the dose limits is part of the optimization, not an optional final detail.

Form the net utility coefficients.
Combine weighted benefit and harm coefficients to obtain $q_0,q_1,q_2$.

Check curvature.
Use the sign of $q_2$ to distinguish a concave maximum, a convex minimum, or a linear/constant function.

Check feasibility and boundaries.
For concave utility, use the stationary dose if it is feasible; otherwise use the appropriate endpoint. For convex or linear utility, compare endpoint utilities.

Worked Example: A Concave Utility Curve with an Interior Optimum

Suppose a hypothetical model gives utility, in utility units, as $U(d)=3+0.8d-0.1d^2$, with dose $d$ in mg and feasible range $0\le d\le5$ mg. Find and verify the maximizing dose.

Here $q_1=0.8$ utility units per mg and $q_2=-0.1$ utility units per mg squared, so the utility is concave. Its stationary dose is $$ d^*=-\frac{0.8}{2(-0.1)}=4\text{ mg}. $$ This lies within the allowed range. Substitution gives $$ U(4)=3+0.8(4)-0.1(4)^2=3+3.2-1.6=4.6 $$ utility units. The endpoint values are $$ U(0)=3,\qquad U(5)=3+0.8(5)-0.1(5)^2=3+4-2.5=4.5. $$ Thus the interior dose of 4 mg gives greater utility than either endpoint and is the maximum on the feasible interval.

Worked Example: Combining Quadratic Benefit and Harm

For a hypothetical treatment, let benefit be $B(d)=2+1.2d-0.12d^2$ benefit units and harm burden be $R(d)=0.2+0.15d+0.03d^2$ burden units, with dose in mg. Use $w_B=1$ utility unit per benefit unit and $w_R=2$ utility units per burden unit. Find the maximum on $0\le d\le4$ mg.

The utility is $U(d)=B(d)-2R(d)$. Combining coefficients gives $$ q_0=2-2(0.2)=1.6,\quad q_1=1.2-2(0.15)=0.9,\quad q_2=-0.12-2(0.03)=-0.18. $$ Therefore $U(d)=1.6+0.9d-0.18d^2$, in utility units, and $q_2<0$. The stationary dose is $$ d^*=-\frac{0.9}{2(-0.18)}=2.5\text{ mg}. $$ It is feasible. Its utility and the endpoint utilities are $$ U(2.5)=1.6+0.9(2.5)-0.18(2.5)^2 =1.6+2.25-1.125=2.725, $$ $$ U(0)=1.6,\qquad U(4)=1.6+0.9(4)-0.18(4)^2=1.6+3.6-2.88=2.32. $$ The 2.5 mg dose is the utility maximum on the range. The harm weight has a visible role: increasing it changes the combined linear and quadratic coefficients, so it can shift the optimum even when the benefit model is unchanged.

Worked Example: A Concave Curve Whose Optimum Is at the Boundary

Suppose a hypothetical model has $U(d)=1+1.1d-0.1d^2$ utility units for doses $0\le d\le4$ mg. Determine the constrained optimum.

The unconstrained stationary dose is $$ d^*=-\frac{1.1}{2(-0.1)}=5.5\text{ mg}. $$ Although this is the unconstrained maximum, it is not feasible. Since $q_2=-0.1<0$ and $d^*>4$, utility is still increasing throughout the allowed range. Compare the feasible endpoints: $$ U(0)=1,\qquad U(4)=1+1.1(4)-0.1(4)^2=1+4.4-1.6=3.8. $$ Thus the largest feasible dose, 4 mg, maximizes utility. The model does not justify recommending 5.5 mg when the permitted range ends at 4 mg.

Interpreting Coefficients and Weights

The maximizing dose depends on the ratio of the net linear and quadratic terms, not on either response curve in isolation. When $q_2<0$, a larger positive $q_1$, holding $q_2$ fixed, moves the stationary dose higher. A more negative $q_2$, holding $q_1>0$ fixed, moves it lower because net utility turns downward more sharply. These statements concern the combined utility coefficients; they do not imply that increasing a benefit-model coefficient always raises the chosen dose, since that change may also alter the curvature or interact with harm.

A simple case illustrates the role of weights. If benefit is increasing and concave while harm is increasing and convex, increasing the harm weight typically makes the combined utility curve bend downward more strongly and may lower the optimal dose. But the precise direction should be verified from the changed coefficients and feasible endpoints. The formula $d^*=-q_1/(2q_2)$ makes clear that a coefficient's effect depends on both numerator and denominator.

A constant term such as $q_0$ changes the utility values but not the stationary dose. It can still matter when reporting absolute utility or comparing utilities across models, provided the utility scale and origin are comparable. By contrast, multiplying the entire utility function by the same positive constant preserves its maximizing dose, while changing the relative benefit and harm weights need not.

Diagnostics, Alternatives, and Limitations

Before using the result, verify that the formula matches the intended utility, the sign convention for harm is correct, weights and units are consistent, and the modeled dose range is the feasible range of interest. Then inspect the fitted or specified response curves over that range. A quadratic can turn sharply or reverse direction outside the data-supported region; an algebraically valid maximum may therefore rely on implausible extrapolation.

Wrong curvature interpretation: A stationary point with $q_2>0$ is a minimum, not a maximum. Check the second derivative or compare utility values.
Ignoring the dose limits: An unconstrained vertex outside the allowable range is not a feasible recommendation. Evaluate the relevant boundary.
Optimizing benefit alone: The vertex of $B(d)$ need not maximize $w_BB(d)-w_RR(d)$. Optimize the combined utility.
Mixing units or signs: Adding unscaled benefit and harm, or treating larger harm as favorable, changes the objective. Define the utility and units before combining coefficients.
Treating a continuous optimum as an available dose: If doses are available only on a grid, evaluate utility at the permitted dose values. The continuous optimum may lie between them.

For a continuous interval and a true quadratic, the derivative-and-boundary method gives an exact answer. A direct endpoint comparison is especially useful for convex, linear, or boundary cases. A numerical grid search is a useful check, and it extends naturally to dose sets with discrete options, but a coarse grid may miss a narrow optimum. For discrete choices, calculate utility at every allowed dose rather than rounding the continuous vertex automatically.

The method is limited by the response model and chosen utility definition. A quadratic is a convenient approximation, not a guarantee that biology follows a parabola. It may not represent plateaus, thresholds, multiple peaks, or asymmetric dose effects well. Moreover, a maximum at a boundary may reflect the restricted dose range rather than a natural optimum. These calculations treat model parameters and preferences as fixed; they do not by themselves quantify uncertainty or ensure safety under a separate hard constraint.

Key takeaway: Combine benefit and harm into $U(d)=q_0+q_1d+q_2d^2$, use curvature to classify any stationary point, and enforce the feasible dose range by checking its boundaries. The vertex is a candidate—not automatically the final dose recommendation.

Check Your Understanding

Work through the questions, then check the concise solutions below.

For $U(d)=2+0.6d-0.15d^2$ utility units on $0\le d\le3$ mg, find the stationary dose and determine whether it is feasible.
For a quadratic with $q_2>0$ on a nondegenerate closed interval, where must a maximum occur? What role does the stationary point play?
A concave utility has stationary dose 7 mg, but the feasible interval is 1 to 5 mg. Which dose maximizes utility, and why?
Suppose benefit and harm are both quadratic. Why is the vertex of the benefit curve alone generally insufficient to select a utility-maximizing dose?

Solutions. (1) $d^*=-0.6/[2(-0.15)]=2$ mg, which is feasible; because $q_2<0$, it is the maximum. (2) A convex quadratic has its maximum at one or both endpoints; its stationary point, if present, is a minimum. Compare the endpoint utilities. (3) The maximum is at 5 mg: the concave curve's vertex lies above the feasible interval, so utility is increasing over the allowed doses. (4) Utility combines the weighted benefit and harm coefficients. Harm can change both the linear term and curvature, shifting the stationary point or making a boundary optimal, so optimize the combined utility instead.