Deriving The First-Order Condition For Optimal Price

Why Differentiate Expected Profit With Respect to Price?

A price change affects both the premium received per policy and the number of policies expected to be purchased. Raising price increases the margin on policies that remain, but may also cause some customers not to buy. An interior optimal price balances these effects. The balance is expressed by a first-order condition, derived by differentiating expected profit with respect to price.

Let $p$ denote the price per policy and let $Q(p)$ denote the expected number of policies purchased at that price. For the basic model, assume that each purchased policy has the same expected cost $c$, independent of price. This cost can include expected claims and other variable costs included in the profit objective. Any fixed cost that does not depend on price changes the level of expected profit but not its derivative.

Definition: In the constant-cost model, expected profit at price $p$ is $\Pi(p)=Q(p)(p-c)$. The function $Q$ is differentiable on the price range under consideration. An interior price is a price inside the feasible range, rather than at a boundary. An interior differentiable maximum must satisfy $\Pi'(p)=0$, but that condition alone does not establish that the price is a maximum.

Deriving the First-Order Condition

Differentiate expected profit using the product rule. The derivative of $Q(p)$ is $Q'(p)$, and the derivative of the per-policy margin $p-c$ is $1$, because $c$ is constant. Therefore,

\Pi'(p)=Q'(p)(p-c)+Q(p).

At an interior optimum $p^*$, the necessary first-order condition is

$$ Q(p^*)+(p^*-c)Q'(p^*)=0. $$

When expected purchases decrease as price increases, $Q'(p^*)<0$. Rearranging the condition gives

Q(p^*)=(p^*-c)\bigl(-Q'(p^*)\bigr).

This form explains the balance. A small price increase adds approximately one dollar of margin on each of the $Q(p^*)$ expected policies. At the same time, it reduces expected purchases by approximately $-Q'(p^*)$ policies per dollar; the margin lost on those policies is approximately $(p^*-c)(-Q'(p^*))$. At the optimum, these effects are equal.

The condition is necessary for an interior optimum under differentiability, not sufficient by itself. A stationary point could instead be a minimum or another type of stationary point. A local maximum can be confirmed, for example, by showing $\Pi''(p^*)<0$. To establish a global optimum, the whole feasible price range and any boundary candidates may also need to be checked.

Elasticity and the Marginal Revenue Interpretation

When $Q(p)>0$ and $Q'(p)<0$, define the positive price elasticity of expected purchases by

\varepsilon(p)=-\frac{pQ'(p)}{Q(p)}.

Dividing the first-order condition by $Q(p^*)$ and using this definition gives

1-\varepsilon(p^*)\frac{p^*-c}{p^*}=0, \qquad \frac{p^*-c}{p^*}=\frac{1}{\varepsilon(p^*)}.

Thus, in the constant-cost model, the optimal margin as a proportion of price equals the reciprocal of the price elasticity at the optimum. This relation requires positive expected purchases, a downward-sloping demand function, and a nonzero price. It does not say that elasticity is constant as price changes; the elasticity in the equation is evaluated at the candidate optimum.

The same condition has the familiar marginal revenue and marginal cost interpretation. If $Q$ can be used locally as the quantity variable, then the inverse demand curve has slope $dp/dQ=1/Q'(p)$. Revenue is $pQ$, so marginal revenue with respect to quantity is

\frac{d(pQ)}{dQ}=p+Q\frac{dp}{dQ} =p+\frac{Q}{Q'(p)}.

Total variable cost is $cQ$, whose marginal cost with respect to quantity is $c$. Setting marginal revenue equal to marginal cost gives $p+Q/Q'=c$, which rearranges to $Q+(p-c)Q'=0$. This interpretation assumes a locally invertible demand curve, in particular $Q'(p)\ne 0$. If demand is not locally invertible, the direct derivative of profit with respect to price remains the appropriate condition.

Worked Example: Checking a Candidate Under Exponential Demand

A hypothetical insurer expects $Q(p)=400e^{-0.02(p-110)}$ policies to be purchased at price $p$, measured in dollars per policy. Expected cost is a constant $c=\$60$ per purchased policy, and feasible prices are nonnegative. Check whether $p=\$110$ satisfies the first-order condition and determine whether it is a maximum. Use differentiation, and avoid the mistake of treating the purchase count as fixed when price changes.

Step 1: Differentiate demand. The chain rule gives $Q'(p)=-0.02Q(p)$. At $p=110$, $Q(110)=400e^0=400$, so $Q'(110)=-0.02(400)=-8$ policies per dollar.

Step 2: Differentiate profit and check the condition. Expected profit is $\Pi(p)=Q(p)(p-60)$, so $\Pi'(p)=Q(p)+(p-60)Q'(p)$. At $p=110$,

\Pi'(110)=400+(110-60)(-8)=400-400=0.

The candidate satisfies the interior first-order condition. Equivalently, the margin is $\$50$, and the purchase response is 8 fewer expected policies per dollar; $50(8)=400$, matching the 400 expected policies whose margin increases by one dollar per dollar of price increase.

Step 3: Check the second derivative. Since $Q'=-0.02Q$, differentiating profit again gives $\Pi''(p)=Q'(p)[1-0.02(p-60)]-0.02Q(p)$. At $p=110$, the bracket is $1-0.02(50)=0$, so $\Pi''(110)=-0.02(400)=-8<0$. The candidate is a strict local maximum. Moreover, $\Pi'(p)=Q(p)[1-0.02(p-60)]$; for nonnegative prices, this is positive below 110 and negative above 110, so this price is also the global maximum on the stated feasible range.

Step 4: Check the profit. Expected profit is $400(110-60)=\$20{,}000$. The calculation is consistent with a positive per-policy margin and 400 expected purchases. The common mistake would be to differentiate only the margin and omit $Q'(p)(p-c)$, thereby ignoring the profit effect of the change in purchases.

When Expected Cost Also Depends on Price

The constant-cost condition relies on expected cost per purchased policy being independent of price. If price changes the mix of customers who purchase, their average expected cost may also change. Let $c(p)$ denote expected cost per purchase at price $p$. Expected profit is then $\Pi(p)=Q(p)[p-c(p)]$, and differentiation gives

\Pi'(p)=Q(p)\bigl[1-c'(p)\bigr]+Q'(p)\bigl[p-c(p)\bigr].

The interior first-order condition is therefore $Q(1-c')+Q'(p-c)=0$, with each function evaluated at the candidate price. The additional term $c'(p)$ matters: a price change can alter expected cost per policy as well as policy count. The constant-cost formula is recovered when $c'(p)=0$. Applying it without checking the cost assumption can give an incorrect result when price-sensitive selection changes the expected cost of the remaining book.

Worked Example: Accounting for Price-Dependent Expected Cost

For a hypothetical product, expected purchases are $Q(p)=600e^{-0.01(p-100)}$, and expected cost per purchased policy is $c(p)=40+0.1(p-100)$, all dollar amounts being per policy. Consider prices from $\$100$ to $\$160$. Find the interior stationary price, verify that it is a maximum, and calculate expected profit there. The method is to differentiate the full profit function, including $c(p)$; the common mistake is to treat the cost as constant.

Step 1: Write the derivatives. Demand satisfies $Q'(p)=-0.01Q(p)$. Cost can be written $c(p)=30+0.1p$, so $c'(p)=0.1$, and the per-policy margin is $p-c(p)=0.9p-30$.

Step 2: Apply the generalized condition. Dividing the derivative of profit by the positive quantity $Q(p)$ gives

\frac{\Pi'(p)}{Q(p)} =1-c'(p)+\frac{Q'(p)}{Q(p)}[p-c(p)] =0.9-0.01(0.9p-30) =1.2-0.009p.

Setting this expression equal to zero gives $p=1.2/0.009=133.333\ldots$. The stationary price is $\$133.33$ when rounded to the nearest cent.

Step 3: Verify the maximum. The derivative has the sign of $1.2-0.009p$, since $Q(p)>0$. It is positive below $133.333\ldots$ and negative above it. Thus profit increases before this price and decreases after it throughout the stated feasible range, confirming a maximum.

Step 4: Calculate the result at the stationary price. At the unrounded price $p=133.333\ldots$, expected cost per policy is $30+0.1p=43.333\ldots$, and the margin is $\$90$. Expected purchases are $600e^{-1/3}$, approximately $429.9188$. Expected profit is therefore $600e^{-1/3}(90)=\$38{,}692.69$, rounded to the nearest cent. The check is that the derivative changes from positive to negative at the calculated price; using the constant-cost condition would omit the $c'(p)=0.1$ term and fail that check.

Using the Condition Carefully

The first-order condition is useful for interpreting and checking a proposed price, but its assumptions should be stated before applying it. In the basic model, price is a continuous decision, expected purchases are differentiable, and expected cost per purchase is constant. If price must be selected from a discrete set, or if it lies at a price floor or ceiling, an interior derivative condition may not determine the best feasible choice. Those cases require comparing feasible choices or checking the relevant boundary.

A practical calculation can proceed in this order:

Specify the profit function.
State whether expected cost per purchase is constant or varies with price, and include any variable costs that belong in the objective.

Differentiate with respect to price.
Apply the product rule to expected purchases and per-policy margin; do not hold purchases fixed unless the model actually makes them independent of price.

Check the candidate and its type.
Set the derivative equal to zero only for an interior candidate, then use a second-derivative or sign check and consider the feasible boundaries.

A useful directional check follows directly from the derivative. Under constant cost, if $Q+(p-c)Q'>0$, a small price increase raises expected profit locally. If $Q+(p-c)Q'<0$, a small price increase lowers expected profit locally. This is a local statement, not a guarantee about a large price change or the global optimum.

Key takeaway: With differentiable expected purchases and constant expected cost per policy, an interior profit-maximizing price satisfies $Q+(p-c)Q'=0$. The added margin on expected purchases balances the margin lost through reduced purchases. If expected cost changes with price, differentiate that dependence as well.

Check Your Understanding

For each problem, use the stated model and check the relevant derivative rather than assuming that a candidate price is optimal.

A hypothetical service has $Q(p)=500-2p$ expected purchases and constant expected cost $c=\$40$. At $p=\$100$, calculate $\Pi'(p)$ and state whether a small price increase raises or lowers expected profit. (Answer: $Q=300$, $\Pi'=300+60(-2)=180>0$; a small price increase raises expected profit.)
Expected purchases for a hypothetical policy are $Q(p)=1000e^{-0.01(p-100)}$, with constant expected cost $c=\$50$. At $p=\$150$, is the price stationary, and is it a local maximum? (Answer: $Q'=-0.01Q$ and $\Pi'=Q[1-0.01(150-50)]=0$; $\Pi''=-0.01Q<0$, so it is a strict local maximum.)
A hypothetical portfolio has $Q(p)=700-2p$ and expected cost per purchase $c(p)=35+0.05(p-100)$. At $p=\$100$, calculate the profit derivative and its sign. (Answer: $Q=500$, $Q'=-2$, $c=35$, and $c'=0.05$; $\Pi'=500(0.95)+(-2)(65)=345>0$.)
Let $Q(p)=400e^{-0.02(p-80)}$ for a hypothetical product with constant expected cost $c=\$30$ and feasible prices $p\geq 0$. Find the stationary price and expected profit there, and state why it is a maximum. (Answer: $Q'=-0.02Q$, so the condition gives $p-30=50$, or $p=\$80$; $Q=400$, expected profit is $\$20{,}000$, and the profit derivative changes from positive to negative there.)
A hypothetical product has $Q(p)=500e^{-0.02(p-100)}$ and $c(p)=30+0.1p$, with feasible prices from $\$100$ to $\$150$. Find the interior stationary price using the price-dependent-cost condition and identify the derivative’s sign on either side. (Answer: $\Pi'/Q=0.9-0.02(0.9p-30)=1.5-0.018p$; the stationary price is $p=83.33\ldots$, outside the feasible range, so there is no interior stationary price on $[\$100,\$150]$; the derivative is negative throughout that range.)

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