Why Reaction Curves Matter
In Cournot competition, firms choose quantities simultaneously, and each firm’s profit depends on the total quantity supplied by all firms. A firm therefore needs to consider not only market demand and its own cost, but also the quantity it expects its rival to produce. A reaction curve makes that strategic dependence visible: it shows the firm’s profit-maximizing quantity for every possible quantity of its rival.
For a two-firm Cournot model, equilibrium can be located graphically. Plot both firms’ reaction curves in a diagram whose axes are the two firms’ quantities. Where the curves intersect, each firm’s quantity is a best response to the other’s. That mutual best response is a Cournot equilibrium, provided the point lies in the feasible quantity region.
Set Up the Quantity-Space Graph
Use \(q_1\) on the horizontal axis and \(q_2\) on the vertical axis. Every point in the graph is a pair of quantities: its horizontal coordinate is firm 1’s output, and its vertical coordinate is firm 2’s output. Quantities cannot be negative, so the economically feasible region is the first quadrant, including its axes.
To make the graph concrete, suppose the firms sell a homogeneous product and face inverse market demand \(P=a-b(q_1+q_2)\), where \(a>0\) is the demand intercept and \(b>0\) measures how much price falls as total output rises. Suppose each firm has constant marginal cost, \(c_1\) and \(c_2\), with no fixed costs. Firms choose nonnegative quantities simultaneously, and each firm takes the rival’s quantity as given when choosing its own. The reaction functions below are for this model; different demand or cost assumptions can produce different curve shapes.
Under these assumptions, firm 1’s profit at a fixed \(q_2\) is \([a-b(q_1+q_2)-c_1]q_1\). Its best response is \(q_1=\max(0,(a-c_1-bq_2)/(2b))\). Firm 2’s profit gives \(q_2=\max(0,(a-c_2-bq_1)/(2b))\). The maximum with zero matters: if the rival produces enough to make any positive output unprofitable, the firm chooses zero rather than a negative quantity.
For graphing, rewrite each positive-output portion with \(q_1\) on the horizontal axis and \(q_2\) on the vertical axis:
This choice of axes is important. Firm 1’s positive-output reaction curve has slope \(-2\) when drawn with \(q_1\) horizontally and \(q_2\) vertically. Firm 2’s curve has slope \(-1/2\) in the same diagram. Both slopes are negative: a greater rival output reduces the output that maximizes a firm’s profit. The two curves need not have the same intercepts or the same feasible endpoints.
Why the Best-Response Formula Has This Form
The formula follows by maximizing each firm’s profit while treating the rival’s quantity as fixed. For firm 1, differentiate profit with respect to \(q_1\). The second derivative is negative, so the positive stationary point, when feasible, is a maximum. If that stationary point is negative, the nonnegativity constraint makes zero the maximizing output.
Setting the first derivative equal to zero gives \(q_1=(a-c_1-bq_2)/(2b)\). Since \(q_1\) must be nonnegative, the best response is the greater of zero and this stationary quantity. Applying the same reasoning to firm 2 gives its corresponding best response. This derivation explains why the curves describe optimal choices, rather than simply combinations of quantities that clear the market.
Worked Example: Finding an Interior Intersection
Worked Example: Finding an Interior Intersection
Two firms face inverse demand \(P=120-2(q_1+q_2)\). Their constant marginal costs are \(c_1=20\) and \(c_2=40\), and there are no fixed costs. Write and graph their reaction curves, locate the equilibrium, and calculate the market price and each firm’s profit.
Method. First write each best response, then express both curves in the chosen graph coordinates. Find their intersection algebraically and verify it by substituting each rival’s output into the other firm’s best response. This avoids treating an approximate visual crossing as an exact answer.
Here \(a=120\), \(b=2\), \(c_1=20\), and \(c_2=40\). Firm 1’s best response is \(q_1=\max(0,(120-20-2q_2)/4)=\max(0,25-0.5q_2)\). Firm 2’s best response is \(q_2=\max(0,(120-40-2q_1)/4)=\max(0,20-0.5q_1)\).
For firm 1’s positive-output portion, the graph equation is \(q_2=50-2q_1\). It crosses the horizontal axis at \((25,0)\) and the vertical axis at \((0,50)\). Firm 2’s curve is \(q_2=20-0.5q_1\), crossing the vertical axis at \((0,20)\) and reaching \(q_2=0\) at \((40,0)\). These intercepts let us sketch both curves before calculating their exact crossing.
At their intersection, both best-response equations hold. Substitute firm 2’s equation into firm 1’s: \(q_1=25-0.5(20-0.5q_1)=25-10+0.25q_1=15+0.25q_1\). Thus \(0.75q_1=15\), so \(q_1=20\). Then \(q_2=20-0.5(20)=10\). Both quantities are positive, so this is an interior intersection.
Check the best responses: against \(q_2=10\), firm 1 chooses \(25-0.5(10)=20\); against \(q_1=20\), firm 2 chooses \(20-0.5(20)=10\). The intersection is therefore a Cournot equilibrium. Total output is \(20+10=30\), so the market price is \(120-2(30)=60\). Firm 1’s profit is \((60-20)(20)=800\), and firm 2’s profit is \((60-40)(10)=200\). The calculated price and profits follow directly from the stated demand and cost assumptions.
Reading the Curves and Their Intersection
The graph gives two related pieces of information. First, each point on firm 1’s reaction curve identifies its optimal \(q_1\) for the rival output \(q_2\) shown at that point. Each point on firm 2’s reaction curve identifies its optimal \(q_2\) for the firm 1 output shown there. Second, the intersection identifies a pair of outputs that satisfies both optimal-choice conditions at once.
An intersection is not merely a place where two lines cross on paper. Its horizontal coordinate must be the quantity firm 1 optimally chooses at the intersection’s vertical coordinate, and its vertical coordinate must be the quantity firm 2 optimally chooses at its horizontal coordinate. These are the two checks that make the point an equilibrium. If either firm could increase its profit by changing its own quantity while the other firm held its quantity fixed, the point would not be a Cournot equilibrium.
For the linear-demand, constant-marginal-cost model, positive-output portions of the reaction curves are straight lines. When a firm’s unconstrained best response falls to zero, its actual best response remains zero. In the chosen graph, this can add a segment on an axis: firm 1’s zero-output responses lie on the vertical axis, while firm 2’s zero-output responses lie on the horizontal axis. Ignoring these boundary segments can hide an equilibrium or lead to an impossible negative quantity.
Worked Example: Locating an Equilibrium on an Axis
Suppose inverse demand is \(P=90-3(q_1+q_2)\), with constant marginal costs \(c_1=30\) and \(c_2=75\), and no fixed costs. Find the reaction curves and equilibrium, then explain why solving the two interior equations without checking feasibility can fail.
Method. Calculate both constrained best responses, identify where each firm’s unconstrained response becomes zero, and test the candidate intersection. This avoids accepting a negative quantity from the interior equations as a feasible solution.
Firm 1’s best response is \(q_1=\max(0,(90-30-3q_2)/6)=\max(0,10-0.5q_2)\). Firm 2’s is \(q_2=\max(0,(90-75-3q_1)/6)=\max(0,2.5-0.5q_1)\). In the graph, firm 1’s positive-output line has intercepts \((10,0)\) and \((0,20)\), followed by a zero-output segment on the vertical axis for \(q_2\geq20\). Firm 2’s positive-output line runs from \((0,2.5)\) to \((5,0)\), followed by a zero-output segment on the horizontal axis for \(q_1\geq5\).
If we set the two positive-output equations equal, substitution gives \(q_1=10-0.5(2.5-0.5q_1)=8.75+0.25q_1\). This implies \(q_1=11.6667\), and then \(q_2=2.5-0.5(11.6667)=-3.3333\). The negative value is not feasible. The calculation signals that the assumed intersection of the two positive-output portions is not an equilibrium.
At \(q_2=0\), firm 1’s best response is \(q_1=10\). At \(q_1=10\), firm 2’s unconstrained response is \(2.5-0.5(10)=-2.5\), so its constrained best response is \(q_2=0\). The curves therefore intersect at \((10,0)\), a boundary equilibrium. Total output is 10 and price is \(90-3(10)=60\). Firm 1 earns \((60-30)(10)=300\); firm 2 earns zero because it produces nothing. The nonnegative-output constraint is essential to this result.
Worked Example: Building a Graph from Intercepts
Worked Example: Building a Graph from Intercepts
Consider inverse demand \(P=84-2(q_1+q_2)\), with constant marginal costs \(c_1=12\) and \(c_2=24\), and no fixed costs. Determine useful intercepts for sketching both reaction curves, calculate their intersection, and find the market price and profits.
Method. Convert the best responses into graph equations, use their intercepts to sketch the lines, and then solve the two response equations exactly. This avoids relying on a drawn intersection to supply precise output values.
Firm 1’s best response is \(q_1=\max(0,(84-12-2q_2)/4)=\max(0,18-0.5q_2)\). Its graph equation is \(q_2=36-2q_1\), with intercepts \((18,0)\) and \((0,36)\). Firm 2’s best response is \(q_2=\max(0,(84-24-2q_1)/4)=\max(0,15-0.5q_1)\), with intercepts \((30,0)\) and \((0,15)\). All four intercepts are nonnegative, so both positive-output portions can be sketched in the first quadrant.
At the crossing, \(q_1=18-0.5q_2\) and \(q_2=15-0.5q_1\). Substitution gives \(q_1=18-0.5(15-0.5q_1)=10.5+0.25q_1\). Therefore \(0.75q_1=10.5\), so \(q_1=14\). Then \(q_2=15-0.5(14)=8\). Checking directly, firm 1 chooses \(18-0.5(8)=14\), and firm 2 chooses \(15-0.5(14)=8\); the crossing is an interior Cournot equilibrium.
Total output is \(14+8=22\), so price is \(84-2(22)=40\). Firm 1’s profit is \((40-12)(14)=28(14)=392\). Firm 2’s profit is \((40-24)(8)=16(8)=128\). The values are consistent with the stated costs and the common market price.
Common Graphing Mistakes
- Swapping the axes without changing the equations. With \(q_1\) horizontal and \(q_2\) vertical, firm 1’s positive-output line is \(q_2=(a-c_1)/b-2q_1\). The original best-response equation has \(q_1\) as a function of \(q_2\); plotting that equation with the axes in the opposite order changes how its slope appears.
- Calling any crossing an equilibrium without checking best responses. Substitute the rival’s quantity into each firm’s response function. Both outputs must be optimal at the same point.
- Extending a line to negative quantities. Quantities are constrained to be nonnegative. Where the unconstrained best response is below zero, the firm’s actual response is zero.
- Reading an approximate graph as an exact calculation. Intercepts and a sketch help locate the crossing, but solve the response equations to obtain exact quantities when possible.
- Confusing a reaction curve with market demand. A reaction curve describes one firm’s optimal output conditional on its rival’s output. Inverse demand describes the market price conditional on total output.
Reaction-curve graphs are a compact way to understand strategic interdependence: moving along one firm’s curve changes the rival output to which its best response applies. The method is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus. In applied work, the assumptions behind the curves matter: the linear equations used here rely on homogeneous-product linear inverse demand, constant marginal costs, and simultaneous quantity choice. A graph based on a different model must use that model’s own best responses.
Check Your Understanding
For each problem, use \(q_1\) as the horizontal axis and \(q_2\) as the vertical axis. Assume the stated linear inverse demand, constant marginal costs, simultaneous quantity choice, nonnegative outputs, and no fixed costs.
- Inverse demand is \(P=100-2(q_1+q_2)\), and marginal costs are \(c_1=20\) and \(c_2=40\). Find both reaction functions, their intersection, and the market price. (Answer: \(q_1=16.6667,\ q_2=6.6667,\ P=53.3333\); the reaction functions are \(q_1=\max(0,20-0.5q_2)\) and \(q_2=\max(0,15-0.5q_1)\).)
- Inverse demand is \(P=72-1.5(q_1+q_2)\), and marginal costs are \(c_1=18\) and \(c_2=36\). Find the intersection and check both best responses. (Answer: \(q_1=16,\ q_2=4\), with each quantity equal to the relevant firm’s best response.)
- Inverse demand is \(P=100-2(q_1+q_2)\), and marginal costs are \(c_1=20\) and \(c_2=80\). Determine whether the equilibrium is interior or on an axis, and give the equilibrium price. (Answer: boundary equilibrium \((q_1,q_2)=(20,0)\), with \(P=60\).)
- Inverse demand is \(P=150-3(q_1+q_2)\), and marginal costs are \(c_1=30\) and \(c_2=60\). Give the two positive-output graph intercepts for each firm’s curve and solve their intersection. (Answer: firm 1 intercepts \((20,0)\) and \((0,40)\); firm 2 intercepts \((30,0)\) and \((0,15)\); intersection \((q_1,q_2)=(16.6667,6.6667)\).)
- Inverse demand is \(P=96-2(q_1+q_2)\), and marginal costs are \(c_1=24\) and \(c_2=48\). Find equilibrium quantities and price, then calculate both profits. (Answer: \(q_1=16,\ q_2=4,\ P=56\); profits are \(\pi_1=512\) and \(\pi_2=32\).)