Why Coupled Demand Changes Pricing
When two products are substitutes, raising the price of one may shift some customers toward the other. A price decision therefore affects not only the product’s own demand, but also the demand for its competitor. The resulting pricing problem is coupled: the best price for either product depends on both prices.
This tutorial models that interaction with linear demand functions and uses multivariable optimization to choose prices that maximize total revenue or profit. A stationary point is not automatically an acceptable pricing decision: the prices must be nonnegative, predicted quantities must be nonnegative, and the objective must have the appropriate curvature.
Linear Demand and Feasible Prices
The negative own-price coefficients mean that increasing a product’s own price reduces its demand, other things equal. The positive cross-price coefficients model substitution: increasing one price raises demand for the other product. The restriction \(C<B\) ensures that the own-price effect dominates the cross-price effect in the symmetric direction where both prices rise together.
Total revenue is the sum of price times quantity for each product: $$ R(p_1,p_2)=p_1q_1+p_2q_2. $$ If each product has the same constant unit cost \(m\ge 0\), total profit is $$ \Pi(p_1,p_2)=(p_1-m)q_1+(p_2-m)q_2. $$ The cost is subtracted from each selling price before multiplying by quantity. Revenue and profit are different objectives, so they need not have the same optimal prices.
Revenue-Maximizing Prices in the Symmetric Model
Substituting the demand equations into revenue gives $$ R=A(p_1+p_2)-B(p_1^2+p_2^2)+2Cp_1p_2. $$ The first-order conditions for an unconstrained stationary point are $$ \frac{\partial R}{\partial p_1}=A-2Bp_1+2Cp_2=0,\qquad \frac{\partial R}{\partial p_2}=A-2Bp_2+2Cp_1=0. $$ Symmetry suggests equal prices, but it is important to establish that the resulting point is a global maximum, not just a symmetric stationary point.
The following change of variables makes the role of equal pricing explicit. Let \(s=p_1+p_2\) be the sum of prices and \(t=p_1-p_2\) their difference. Since \(p_1^2+p_2^2=(s^2+t^2)/2\) and \(2p_1p_2=(s^2-t^2)/2\), revenue becomes $$ R=As-\frac{B-C}{2}s^2-\frac{B+C}{2}t^2. $$ The conditions \(B-C>0\) and \(B+C>0\) make both quadratic penalties strictly positive away from their respective maximizing values. Completing the square gives $$ R=\frac{A^2}{2(B-C)} -\frac{B-C}{2}\left(s-\frac{A}{B-C}\right)^2 -\frac{B+C}{2}t^2. $$ Therefore revenue is largest uniquely when \(s=A/(B-C)\) and \(t=0\). These conditions imply \(p_1=p_2=A/[2(B-C)]\). At those prices, $$ q_1=A-(B-C)p_1=\frac{A}{2},\qquad q_2=A-(B-C)p_2=\frac{A}{2}. $$ The prices and quantities are nonnegative when \(A\ge0\), so the maximizing point is feasible. The completed-square expression also proves it is the unique global maximum, rather than merely a solution of the first-order equations.
Worked Example: Symmetric Revenue Pricing
Worked Example: Symmetric Revenue Pricing
Suppose \(A=24\), \(B=5\), and \(C=2\). Then $$ q_1=24-5p_1+2p_2,\qquad q_2=24-5p_2+2p_1. $$ Because \(5>2\ge0\), the theorem applies. The maximizing prices are $$ p_1=p_2=\frac{24}{2(5-2)}=\frac{24}{6}=4. $$ Substituting these prices into both demand equations gives $$ q_1=24-5(4)+2(4)=24-20+8=12, $$ $$ q_2=24-5(4)+2(4)=24-20+8=12. $$ Both prices and quantities are nonnegative. The total revenue is $$ R=4(12)+4(12)=48+48=96. $$ As a check, the maximum-revenue formula gives $$ \frac{A^2}{2(B-C)}=\frac{24^2}{2(3)}=\frac{576}{6}=96, $$ in agreement with the direct calculation.
How Coupling Changes the Price
With no cross-price effect, \(C=0\), each product’s demand depends only on its own price. The symmetric revenue-maximizing price is then \(A/(2B)\). When \(0<C<B\), the coupled price is \(A/[2(B-C)]\). If \(A>0\), then \(B-C<B\), so $$ \frac{A}{2(B-C)}>\frac{A}{2B}. $$ The condition \(A>0\) matters: when \(A=0\), both expressions are zero, so they are equal rather than strictly ordered.
In this symmetric model, the optimal quantity of each product remains \(A/2\) as \(C\) changes, while the price rises as \(C\) increases (with \(A\) and \(B\) fixed and \(0\le C<B\)). Cross-price responsiveness means a price increase in one product also raises demand for the other. The joint optimum accounts for that interaction; it is not found by maximizing each product’s revenue as though the other price had no effect.
Worked Example: Comparing Coupled and Uncoupled Prices
Worked Example: Comparing Coupled and Uncoupled Prices
Let \(A=18\) and \(B=4\). Without cross-price effects, \(C=0\), so each revenue-maximizing price is $$ p=\frac{18}{2(4)}=\frac{18}{8}=\frac{9}{4}. $$ Now let \(C=1\). Since \(0<1<4\), the coupled optimum is $$ p=\frac{18}{2(4-1)}=\frac{18}{6}=3. $$ The increase is \(3-9/4=12/4-9/4=3/4\). For \(A>0\) and \(0<C<B\), the formula ensures the coupled price exceeds the uncoupled price. At the coupled optimum, each quantity is $$ q_i=\frac{A}{2}=\frac{18}{2}=9. $$ The coupled total revenue is \(3(9)+3(9)=54\); without coupling, each quantity is also \(9\), giving total revenue \(2(9/4)(9)=162/4=81/2\). The calculations confirm that in this example coupling raises both the optimal price and the maximum total revenue.
Unequal Products: Solving the Coupled First-Order Conditions
The symmetric formula depends on identical demand intercepts and identical own-price effects. For unequal products, a useful general model is $$ q_1=a_1-b_1p_1+c_{12}p_2,\qquad q_2=a_2-b_2p_2+c_{21}p_1. $$ Assume \(b_1,b_2>0\) and \(c_{12},c_{21}\ge0\). The two cross-price coefficients need not be equal. Expanding \(R=p_1q_1+p_2q_2\) and differentiating gives $$ \frac{\partial R}{\partial p_1}=a_1-2b_1p_1+(c_{12}+c_{21})p_2, $$ $$ \frac{\partial R}{\partial p_2}=a_2-2b_2p_2+(c_{12}+c_{21})p_1. $$ Thus the stationary prices solve a coupled linear system. A sufficient and exact condition for the revenue Hessian to be negative definite in this model is \(4b_1b_2>(c_{12}+c_{21})^2\). Under this condition, a stationary point is the unique unconstrained global maximum. It is the constrained pricing optimum as well if its prices and predicted quantities are feasible.
Worked Example: Unequal Demand Intercepts
Worked Example: Unequal Demand Intercepts
Consider $$ q_1=30-3p_1+p_2,\qquad q_2=18-2p_2+\frac12p_1. $$ Here \(a_1=30\), \(a_2=18\), \(b_1=3\), \(b_2=2\), and \(c_{12}+c_{21}=1+1/2=3/2\). The first-order equations are $$ 6p_1-\frac32p_2=30,\qquad -\frac32p_1+4p_2=18. $$ The Hessian condition holds because $$ 4b_1b_2=4(3)(2)=24>\left(\frac32\right)^2=\frac94. $$ The determinant of the coefficient matrix is $$ 6(4)-\left(\frac32\right)^2=24-\frac94=\frac{87}{4}. $$ Solving the system gives $$ p_1=\frac{4(30)+(3/2)(18)}{87/4} =\frac{120+27}{87/4} =\frac{588}{87} =\frac{196}{29}, $$ $$ p_2=\frac{(3/2)(30)+6(18)}{87/4} =\frac{45+108}{87/4} =\frac{612}{87} =\frac{204}{29}. $$ The prices are positive. Substituting them into demand, $$ q_1=30-\frac{588}{29}+\frac{204}{29} =\frac{870-588+204}{29} =\frac{486}{29}, $$ $$ q_2=18-\frac{408}{29}+\frac{98}{29} =\frac{522-408+98}{29} =\frac{212}{29}. $$ Both quantities are positive, so the prices are feasible. Substitution into the first stationarity equation gives \(6(196/29)-(3/2)(204/29)=(1176-306)/29=30\); into the second it gives \(-(3/2)(196/29)+4(204/29)=(-294+816)/29=18\). The equations are satisfied, and negative definiteness establishes that this feasible stationary point is the unique global revenue maximum.
Including Unit Costs
For the symmetric model with equal unit cost \(m\ge0\), profit expands to $$ \Pi=A(p_1+p_2)-B(p_1^2+p_2^2)+2Cp_1p_2 -m\bigl(2A-(B-C)(p_1+p_2)\bigr). $$ Its quadratic part is the same as revenue’s, so it is strictly concave when \(B>C\ge0\). Solving the first-order conditions yields equal prices $$ p_1=p_2=\frac{A+(B-C)m}{2(B-C)}. $$ The corresponding quantities are $$ q_1=q_2=A-(B-C)p_1=\frac{A-(B-C)m}{2}. $$ Consequently, this interior profit solution is feasible with positive quantities when \(A>(B-C)m\). If that condition fails, the stationary point does not give a feasible positive-quantity solution; a constrained boundary analysis is needed rather than accepting the formula without checking demand.
Worked Example: Profit With Coupled Demand
Worked Example: Profit With Coupled Demand
Let \(A=20\), \(B=4\), \(C=1\), and \(m=2\). Here \(B-C=3\), and \(A=20>3(2)=6\), so the interior solution has positive quantities. The optimal price of each product is $$ p_i=\frac{20+3(2)}{2(3)}=\frac{26}{6}=\frac{13}{3}. $$ Each quantity is $$ q_i=20-4\left(\frac{13}{3}\right)+\frac{13}{3} =20-\frac{52}{3}+\frac{13}{3} =\frac{60-52+13}{3}=7. $$ Each unit margin is \(13/3-2=7/3\). Total profit is therefore $$ \Pi=2\left(\frac73\right)(7)=\frac{98}{3}. $$ As a check, the first-order condition for either price at equal prices is \(A+(B-C)m-2(B-C)p=0\). Substitution gives \(20+6-6(13/3)=26-26=0\), as required. The computed quantities are positive, confirming feasibility.
Common Pitfall: Ignoring Cross-Price Terms
A common error is to differentiate \(p_1q_1\) as if \(q_1\) depended only on \(p_1\). In the symmetric model, the incorrect calculation would be $$ \frac{\partial (p_1q_1)}{\partial p_1}=q_1+p_1\frac{\partial q_1}{\partial p_1}, $$ and then to treat this as the derivative of total revenue. But total revenue also includes \(p_2q_2\), and \(q_2\) depends on \(p_1\). The correct derivative is $$ \frac{\partial R}{\partial p_1} =q_1+p_1\frac{\partial q_1}{\partial p_1} +p_2\frac{\partial q_2}{\partial p_1} =q_1-Bp_1+Cp_2. $$ Substituting \(q_1=A-Bp_1+Cp_2\) gives \(A-2Bp_1+2Cp_2\), including the effect of the price change on both products’ revenue. The correction is to differentiate the entire objective, not just the revenue term for the product whose price is changing.
Check Your Understanding
Use the formulas and feasibility checks from this tutorial to answer the following questions.
- For \(A=16\), \(B=5\), and \(C=1\), find both revenue-maximizing prices and quantities, then calculate maximum total revenue.
- Why does \(B>C\) matter for the symmetric revenue maximum?
- For positive \(A\) and \(0<C<B\), compare the coupled and uncoupled symmetric revenue prices. What changes if \(A=0\)?
- In the unequal-demand model, which terms appear in \(\partial R/\partial p_1\) because \(p_1\) affects the second product’s demand?
- For the profit model with \(A=15\), \(B=4\), \(C=1\), and \(m=2\), determine whether the interior solution has positive quantities.