Why Nonlinear Monopoly Choice Needs Care
When demand is nonlinear, the marginal revenue curve is generally not a straight line. If costs are nonlinear too, marginal cost may also change at a nonconstant rate. The monopolist still chooses output to maximize profit, but finding the answer requires deriving both marginal functions, solving their equality, and checking that the solution lies in the economically feasible range.
The method in this tutorial is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus. It applies to a single-price monopolist: the firm chooses one quantity, and the inverse-demand curve gives the price buyers will pay for that quantity.
The word interior matters. It means the candidate output is strictly inside the allowed output range, rather than at a boundary such as zero output or the largest quantity buyers will purchase at a nonnegative price. The equality \(MR=MC\) identifies an interior stationary point; by itself, it does not prove that the point is feasible or a maximum.
Deriving Marginal Revenue and the Choice Condition
With nonlinear inverse demand, do not assume that marginal revenue equals price. Revenue changes both because the firm sells one more unit and because selling more generally requires a change in the price charged on all units. The product rule captures both effects.
For a downward-sloping demand curve, \(dP/dQ<0\), so the second term reduces marginal revenue below price whenever \(Q>0\). Profit is \(\pi(Q)=P(Q)Q-C(Q)\). Differentiating gives \(\pi'(Q)=MR(Q)-MC(Q)\). At an interior maximum, a small increase or decrease in output must not raise profit, so the first derivative is zero: \(MR(Q)=MC(Q)\).
A useful local check is the second derivative. Since \(\pi''(Q)=MR'(Q)-MC'(Q)\), a negative value at a stationary point is a sufficient local test for a maximum when the functions are twice differentiable. If \(\pi''(Q)<0\) throughout the feasible interval, profit is strictly concave there, and an interior stationary point is the unique global maximum on that interval. Otherwise, inspect the relevant feasible boundaries and any other stationary points as well.
Worked Nonlinear Monopoly Problems
Worked Example: Quadratic Demand With Quadratic Cost
A hypothetical sole supplier faces inverse demand \(P(Q)=100-2Q-0.1Q^2\) and has total cost \(C(Q)=20+10Q+Q^2\). Find the profit-maximizing output, its price, and profit. Quantities and monetary values are measured in compatible units.
Method. Differentiate total revenue and total cost, solve the marginal condition, and check curvature and demand feasibility. This avoids treating the price on the last unit as marginal revenue.
First, total revenue is \(TR(Q)=(100-2Q-0.1Q^2)Q=100Q-2Q^2-0.1Q^3\). Therefore, \(MR(Q)=100-4Q-0.3Q^2\). Differentiating cost gives \(MC(Q)=10+2Q\). Set them equal:
The two roots are \(Q=10\) and \(Q=-30\). Negative output is infeasible, so the candidate is \(Q=10\). At that output, the price is \(P(10)=100-2(10)-0.1(10^2)=100-20-10=70\). Revenue is \(70(10)=700\). Cost is \(C(10)=20+10(10)+10^2=20+100+100=220\), so profit is \(700-220=480\).
For the maximum check, \(\pi''(Q)=MR'(Q)-MC'(Q)=(-4-0.6Q)-2=-6-0.6Q\), which is negative for every \(Q\geq0\). Demand is nonnegative at \(Q=10\), since the price is 70. Profit is strictly concave over the feasible nonnegative-output range, so this feasible stationary point is the unique maximum there. Answer: output 10, price 70, and profit 480 monetary units.
Worked Example: Squared-Quantity Demand and Rising Marginal Cost
A hypothetical digital service has inverse demand \(P(Q)=100-Q^2\) and total cost \(C(Q)=5Q+2Q^2\). Find the monopoly output and price, and establish that the candidate is a maximum.
Method. Use the product rule for revenue, then check that the resulting root is feasible and that profit is concave. A trap here is to differentiate inverse demand alone and call the result marginal revenue.
Total revenue is \(TR(Q)=(100-Q^2)Q=100Q-Q^3\), so \(MR(Q)=100-3Q^2\). Marginal cost is \(MC(Q)=5+4Q\). Equating these gives
The roots are \(Q=5\) and \(Q=-19/3\). Discard the negative root. At \(Q=5\), price is \(P(5)=100-5^2=75\). Revenue is \(75(5)=375\). Cost is \(C(5)=5(5)+2(5^2)=25+50=75\), giving profit \(375-75=300\).
The demand price is nonnegative for \(0\leq Q\leq10\), so \(Q=5\) is within the demand range. Also, \(\pi''(Q)=(-6Q)-4=-6Q-4<0\) for all \(Q\geq0\). Thus profit is strictly concave and \(Q=5\) is the unique feasible maximum. Answer: output 5, price 75, and profit 300 monetary units.
Worked Example: Nonlinear Demand and Cubic Cost
A hypothetical materials supplier faces \(P(Q)=135-2Q-Q^2\) and has total cost \(C(Q)=5Q+Q^2+Q^3/3\). Find the profit-maximizing output and price, and calculate profit to the nearest cent.
Method. Derive both marginal functions and use curvature to establish the maximum. This example also checks that the price is taken from inverse demand after solving for quantity, rather than being confused with marginal revenue.
Revenue is \(TR(Q)=(135-2Q-Q^2)Q=135Q-2Q^2-Q^3\). Therefore \(MR(Q)=135-4Q-3Q^2\). Differentiating the cost function gives \(MC(Q)=5+2Q+Q^2\). The first-order condition becomes
The roots are \(Q=5\) and \(Q=-6.5\); only \(Q=5\) is feasible. The price is \(P(5)=135-2(5)-5^2=135-10-25=100\). Revenue is \(100(5)=500\). Cost is \(C(5)=5(5)+5^2+5^3/3=25+25+125/3=50+41.666\ldots=91.666\ldots\). Thus profit is \(500-91.666\ldots=408.333\ldots\), or 408.33 monetary units to the nearest cent.
For the second-order check, \(\pi''(Q)=MR'(Q)-MC'(Q)=(-4-6Q)-(2+2Q)=-6-8Q\), which is negative for \(Q\geq0\). The price at the candidate is positive, so it lies within the nonnegative-price demand range. Strict concavity establishes that the feasible stationary point is the unique maximum. Answer: output 5, price 100, and profit 408.33 monetary units.
Feasibility and the Checks That Matter
A polynomial equation can produce roots that have no economic meaning for the problem. A candidate output must be nonnegative and within the range where the stated demand function applies. If the model requires a nonnegative price, also check \(P(Q)\geq0\). For example, a root beyond the quantity at which price reaches zero is not an acceptable monopoly choice under that restriction.
When more than one feasible stationary point exists, compare the associated profit levels and check the feasible boundaries. When the profit function is strictly concave over the entire feasible interval, there can be at most one stationary point, and it is the global maximum if it is feasible and interior. The condition \(MR=MC\) is a candidate-finding rule, not a substitute for these checks.
Use \(TR(Q)=P(Q)Q\) and \(\pi(Q)=TR(Q)-C(Q)\), keeping the given demand and cost functions intact.
Obtain \(MR(Q)=P(Q)+QP'(Q)\) and \(MC(Q)=C'(Q)\). Show the derivatives before solving.
Set \(MR(Q)=MC(Q)\), solve the equation, and discard roots outside the feasible output range.
Substitute the retained quantity into \(P(Q)\). Check price feasibility and use curvature or profit comparisons to confirm a maximum.
Common Mistake: Using Price as Marginal Revenue
Suppose inverse demand is \(P(Q)=80-2Q\). A mistaken calculation may differentiate demand and report \(MR=-2\). That is the slope of the inverse-demand curve, not marginal revenue. Another incorrect shortcut is to set \(P=MC\), which is not the interior monopoly profit condition.
The correction is to calculate total revenue first: \(TR(Q)=(80-2Q)Q=80Q-2Q^2\). Differentiating gives \(MR(Q)=80-4Q\), not \(-2\). If marginal cost were \(MC(Q)=8+Q\), the monopoly condition would be \(80-4Q=8+Q\), so \(5Q=72\) and \(Q=14.4\). Price must then be recovered from demand: \(P(14.4)=80-2(14.4)=51.2\). Substituting this quantity back into the marginal functions checks the result: \(MR(14.4)=80-57.6=22.4\) and \(MC(14.4)=8+14.4=22.4\). The calculation uses marginal revenue, not the demand slope or price, to determine output.
Check Your Understanding
For each problem, derive the relevant marginal functions, solve for the feasible output, and apply the stated checks.
- A hypothetical firm faces \(P(Q)=60-Q^2\) and \(C(Q)=4Q+Q^2\). Find the interior monopoly output and price. (Answer: \(Q=4\), \(P=44\).)
- A producer has inverse demand \(P(Q)=90-2Q^2\) and cost \(C(Q)=12Q+4Q^2\). Find the feasible output and price. (Answer: \(Q=3\), \(P=72\).)
- A supplier faces \(P(Q)=80-Q-0.1Q^2\) and \(C(Q)=10Q+Q^2\). Solve the marginal condition, report the feasible output and price, and check that profit is concave. (Answer: \(Q=10\), \(P=60\); \(\pi''(Q)=-4-0.6Q<0\) for \(Q\geq0\).)
- A hypothetical firm has \(P(Q)=135-2Q-Q^2\) and \(C(Q)=5Q+Q^2+Q^3/3\). Find output, price, and profit to the nearest cent, and verify the maximum using the second derivative. (Answer: \(Q=5\), \(P=100\), profit \(=408.33\); \(\pi''(Q)=-6-8Q<0\) for \(Q\geq0\).)