From a Demand Model to a Profit Model
A firm choosing a price faces a linked change in demand: a higher price generally means fewer units are sold. To maximize profit, the firm must account for both effects. The revenue from a price is not simply that price; it is the price multiplied by the quantity customers will buy at that price.
Suppose demand is described by the linear model \(q(p)=a-bp\), where \(p\) is price, \(q(p)\) is quantity demanded, and \(a\) and \(b\) are positive constants. The model predicts nonnegative sales only when \(0\le p\le a/b\). A cost function \(C(q)\) gives the total cost of producing quantity \(q\). In this tutorial, assume \(C\) is continuous on the quantities considered and differentiable in their interior.
There are two equivalent ways to optimize this profit. We can use price as the variable, or express price in terms of quantity and optimize over quantities. Which form is simpler depends on the cost function and any additional restrictions.
Writing Profit in Terms of Quantity
Solving \(q=a-bp\) for price gives the inverse demand relation \(p=(a-q)/b\). Substituting it into revenue gives \(R(q)=q(a-q)/b\), so profit as a function of quantity is $$ \Pi(q)=\frac{q(a-q)}{b}-C(q),\qquad 0\le q\le a. $$ This representation makes the feasible interval explicit. It also makes it straightforward to check a candidate quantity against the demand model and then recover the corresponding price.
Worked Example: Finding an Interior Price and Checking It
Suppose demand is \(q=120-2p\), and total cost is \(C(q)=300+10q+\tfrac12q^2\). Nonnegative price and quantity require \(0\le p\le60\), or equivalently \(0\le q\le120\).
First write inverse demand: $$ p=\frac{120-q}{2}=60-\frac{q}{2}. $$ Revenue in terms of quantity is $$ R(q)=q\left(60-\frac{q}{2}\right)=60q-\frac12q^2. $$ Subtracting cost gives $$ \Pi(q)=60q-\frac12q^2-\left(300+10q+\frac12q^2\right) =-q^2+50q-300. $$ Differentiate and set the derivative equal to zero: $$ \Pi'(q)=-2q+50=0 \quad\Longrightarrow\quad q=25. $$ This candidate lies in the feasible interval. The endpoint profits and candidate profit are $$ \Pi(0)=-300,\qquad \Pi(120)=-120^2+50(120)-300=-8700, $$ $$ \Pi(25)=-25^2+50(25)-300=325. $$ The candidate therefore gives the largest profit among the endpoints and critical points.
Recover the price and check the original model: $$ p=60-\frac{25}{2}=\frac{95}{2}=47.5, \qquad q=120-2(47.5)=25. $$ At this price, revenue is \(47.5(25)=1187.5\), and cost is $$ C(25)=300+10(25)+\frac12(25^2)=300+250+312.5=862.5. $$ Thus profit is \(1187.5-862.5=325\), matching the quantity calculation.
As a further check, direct substitution into the price form gives $$ \Pi(p)=p(120-2p)-C(120-2p)=-4p^2+380p-8700. $$ Its derivative is \(-8p+380\), which is zero at \(p=47.5\), the same price found from quantity.
Why the Two Optimization Variables Agree
The price and quantity formulations are not two different economic models. The inverse demand relation pairs every feasible price with exactly one feasible quantity, and vice versa, because \(b>0\). Substitution preserves the profit associated with each pair. Consequently, a quantity that maximizes the quantity formula corresponds to a price that maximizes the price formula, provided both are optimized over corresponding feasible intervals.
The derivative condition also has an economic interpretation. If \(C\) is differentiable, differentiating the quantity formula at an interior optimum gives $$ \Pi'(q)=\frac{a-2q}{b}-C'(q)=0. $$ Inverse demand gives \(p=(a-q)/b\), so \((a-2q)/b=p-q/b\). The change in revenue per extra unit is therefore \(p-q/b\), while the change in cost per extra unit is \(C'(q)\). At an interior optimum these are equal.
This is often summarized as marginal revenue equals marginal cost. It is a useful way to locate possible interior optima, not a substitute for checking feasibility or comparing profits. If the cost function is not differentiable at some feasible point, that point must also be considered when searching for a global maximum.
Comparing All Feasible Candidates
For a continuous profit function on a closed feasible interval, a global maximum is attained. If the profit function is differentiable in the interval’s interior, an interior maximum must occur at a critical point, where the derivative is zero or is undefined. Therefore, compare the profit at the endpoints and at all relevant interior critical points. A derivative equation alone identifies candidates; the comparison determines which candidate is best.
Real restrictions can narrow the interval. For example, a permitted price range may exclude some prices even though demand remains nonnegative there. In that case, convert the restricted price interval to its corresponding quantity interval before optimizing in quantity. A critical point outside that interval is not feasible and cannot be the answer.
Worked Example: A Fixed Cost and an Interior Optimum
Suppose \(q=90-3p\) and \(C(q)=20+8q+q^2\). The feasible price interval is \(0\le p\le30\), corresponding to \(0\le q\le90\). Inverse demand is $$ p=\frac{90-q}{3}=30-\frac{q}{3}. $$ Revenue is \(R(q)=30q-\tfrac13q^2\), so profit is $$ \Pi(q)=30q-\frac13q^2-(20+8q+q^2) =-\frac43q^2+22q-20. $$ The derivative and stationary quantity are $$ \Pi'(q)=-\frac83q+22=0 \quad\Longrightarrow\quad q=\frac{66}{8}=\frac{33}{4}. $$ This is feasible because \(0\le33/4\le90\). Compare it with both endpoints: $$ \Pi(0)=-20,\qquad \Pi(90)=-\frac43(90)^2+22(90)-20=-8840, $$ $$ \Pi\left(\frac{33}{4}\right) =-\frac{363}{4}+\frac{726}{4}-\frac{80}{4} =\frac{283}{4}=70.75. $$ Thus the interior candidate is the global maximum on the feasible interval.
Its price is $$ p=30-\frac{1}{3}\left(\frac{33}{4}\right) =30-\frac{11}{4}=\frac{109}{4}=27.25. $$ Check demand: \(90-3(109/4)=33/4\). At this quantity, cost is $$ C\left(\frac{33}{4}\right)=20+66+\frac{1089}{16} =\frac{2465}{16}, $$ and revenue is $$ \frac{109}{4}\cdot\frac{33}{4}=\frac{3597}{16}. $$ Their difference is \((3597-2465)/16=1132/16=283/4\), agreeing with the optimized profit.
When a Constraint Pushes the Best Choice to an Endpoint
Suppose a price restriction changes the feasible interval. An unconstrained critical point may then be irrelevant because it lies outside the permitted choices. The best feasible price can instead occur at a boundary. The following example shows why the interval must be applied before declaring an optimum.
Worked Example: A Price Cap Makes the Boundary Optimal
Let demand be \(q=80-2p\), cost be \(C(q)=25q\), and suppose the permitted price is \(0\le p\le30\). The demand model alone allows prices up to \(40\), but the cap reduces the feasible price interval. Since \(q=80-2p\), the corresponding quantities satisfy \(20\le q\le80\). Inverse demand is \(p=40-q/2\).
Profit in terms of quantity is $$ \Pi(q)=q\left(40-\frac{q}{2}\right)-25q =-\frac12q^2+15q,\qquad 20\le q\le80. $$ The derivative is \(\Pi'(q)=15-q\), whose only zero is \(q=15\). That point is outside the feasible interval. For every feasible \(q\ge20\), the derivative \(15-q\) is negative, so profit decreases throughout the permitted interval. The endpoint values are $$ \Pi(20)=-\frac12(20)^2+15(20)=100, \qquad \Pi(80)=-\frac12(80)^2+15(80)=-2000. $$ Hence the maximum over permitted operating choices occurs at \(q=20\), corresponding to \(p=40-20/2=30\). Checking the original model, demand at price \(30\) is \(80-2(30)=20\), revenue is \(30(20)=600\), cost is \(25(20)=500\), and profit is \(600-500=100\).
This example finds the best choice within the stated operating interval. If the firm is also allowed to stop operating and thereby earn zero profit, it should compare that option with the best operating profit. Here the operating profit is \(100\), so operating remains preferable.
A Common Mistake: Maximizing Revenue Instead of Profit
Revenue and profit are different objectives. Revenue is price times quantity; profit subtracts the cost of producing that quantity. The price that maximizes revenue need not maximize profit.
A Reliable Procedure
For a linear demand and cost problem, keep the economic model and the calculus steps connected:
Identify the demand equation and every price or quantity limit. Use them to determine the feasible interval.
Compute revenue as price times demanded quantity, then subtract the cost of producing that quantity.
Use price directly, or solve demand for price and express profit in quantity. Keep the corresponding feasible interval with the formula.
Check feasible interior critical points and the endpoints. Do not treat a stationary point outside the interval as a possible choice.
Convert the winning quantity back to price, or vice versa, and substitute into the original demand and cost model to check profit.
The key modeling step is to make the quantity sold depend on the chosen price before calculating revenue and cost. Once the profit function and its feasible interval are correct, calculus can identify candidates; endpoint comparisons and a final check establish that the reported choice is feasible and achieves the claimed profit.
Check Your Understanding
Use the demand, cost, and restrictions stated in each question. Show how the feasible interval affects your conclusion.
- If \(q=a-bp\) with \(a,b>0\), what price interval corresponds to nonnegative prices and quantities, and what is the inverse demand relation?
- Why does maximizing revenue alone generally fail to identify the price that maximizes profit?
- State the derivative condition for an interior profit maximum when profit is written as a differentiable function of quantity. Why is the condition not sufficient on its own to establish a global maximum?
- If a price restriction makes the quantity corresponding to a stationary point infeasible, what candidates should be compared on the restricted interval?
- For the price-cap example, verify the maximum profit using price rather than quantity.
