Why Boundaries Matter
A firm’s profit-maximizing output may be an interior quantity, but it may instead occur at a boundary of the quantities the firm can produce. Zero output is one boundary; a production-capacity limit can be another. A derivative condition can identify interior candidates, but it does not by itself decide between those candidates and the feasible endpoints.
The task is to find the global maximum: the feasible output that gives at least as much profit as every other feasible output. This requires comparing profit levels, not merely identifying where marginal revenue equals marginal cost. The earlier material on profit functions and concavity provides useful tools, but the comparison with the boundaries remains necessary when output is constrained.
The value \(\pi(0)\) need not be zero. If costs are incurred even when output is zero, then \(TC(0)>0\), and therefore \(\pi(0)=TR(0)-TC(0)\) may be negative. The zero-output choice must be evaluated using the profit function just like any other feasible quantity.
The Candidate-Comparison Method
Assume profit is continuous on the feasible interval and differentiable at every interior point. Continuity ensures that a maximum exists on the closed, bounded interval. If a maximum occurs strictly between the endpoints, the derivative of profit there must be zero. These facts reduce the search for a global maximum to a finite candidate comparison when the derivative equation has finitely many solutions.
Because \(\pi'(q)=MR(q)-MC(q)\) when the revenue and cost functions are differentiable, an interior stationary point satisfies \(MR(q)=MC(q)\). That equality supplies a candidate, not a global answer. In particular, the stationary point may be a minimum, may be outside the feasible interval, or may yield less profit than an endpoint.
Record the output restriction, usually \(0\le q\le\bar q\). Do not retain a stationary solution that lies outside this interval.
Differentiate \(\pi(q)\) and solve \(\pi'(q)=0\), keeping only solutions strictly inside the feasible interval.
Calculate profit at each retained stationary point and at both endpoints, including \(q=0\).
The candidate with the largest profit is a global maximizer. If several candidates tie for the largest value, each is a global maximizer.
Why the Candidate List Is Complete
Here is the justification for comparing endpoints and stationary points. Since \(\pi\) is continuous on the closed interval \([0,\bar q]\), the extreme value theorem guarantees that it attains a highest value somewhere on that interval. Let \(q^*\) be a point where that highest value is attained.
If \(q^*=0\) or \(q^*=\bar q\), it is an endpoint candidate. Otherwise, \(q^*\) is an interior point. At an interior maximum of a differentiable function, small movements either side cannot increase the function. The difference quotients from the right are therefore nonpositive near the maximum, while those from the left are nonnegative. If the derivative exists, the two one-sided limits must agree, so their common value is zero: \(\pi'(q^*)=0\). Thus any global maximum must be an endpoint or an interior stationary point. Comparing the profit values at all such candidates identifies the maximum.
This argument relies on the stated assumptions. If profit is discontinuous or the feasible set is not a closed, bounded interval, a maximum is not guaranteed by this result. If profit has a kink where it is not differentiable, that feasible kink must also be checked as a candidate; the derivative-zero test applies only at differentiable interior points.
Worked Example: An Interior Maximum Ties with Zero Output
A hypothetical producer can choose any output from zero to 4 units. Its total revenue and total cost, in dollars, are \(TR(q)=15q\) and \(TC(q)=q^3-6q^2+24q+2\). Find every global profit-maximizing output.
Method. Construct the profit function, find all feasible interior stationary points, and compare profit at those points and at both endpoints. This method avoids assuming that an interior stationary point is the unique best choice.
First calculate profit:
Differentiate and factor:
The stationary outputs are \(q=1\) and \(q=3\), and both are feasible because each lies between 0 and 4. The complete candidate list is therefore \(q=0,1,3,4\). Evaluate each:
The largest profit is \(-2\), attained at both \(q=0\) and \(q=3\). Thus there are two global maximizers. The result also illustrates that a global maximum can be negative: it is the highest profit among the feasible choices, even when every candidate produces a loss.
Check. The candidate list includes both endpoints and both feasible stationary points, and neither \(q=1\) nor \(q=4\) has profit as high as \(-2\). Considering only the interior stationary point with the highest value would miss the tie with zero output.
Worked Example: The Capacity Boundary Is Best
A hypothetical workshop can make between zero and 2 units. Its revenue is \(TR(q)=18q\) dollars and its cost is \(TC(q)=2q^2+6q+10\) dollars. Determine the profit-maximizing feasible output.
Method. Solve the stationary condition, test whether the solution is feasible, and then compare the endpoint profits. This avoids treating an unconstrained stationary solution as feasible when it exceeds capacity.
The profit function and its derivative are:
Setting the derivative equal to zero gives \(-4q+12=0\), so the stationary output is \(q=3\). But the feasible interval is \([0,2]\), and \(3\) is not feasible. There is no interior stationary point in the interval. The candidates are just \(q=0\) and \(q=2\):
Therefore, the global maximum on the feasible interval is 6 dollars at \(q=2\). The stationary solution \(q=3\) cannot be selected, regardless of what profit the formula assigns to that infeasible quantity.
Check. At the capacity boundary, \(\pi'(2)=-4(2)+12=4>0\). Profit is still increasing as output approaches the boundary from within the feasible interval, consistent with the boundary being better than the zero-output endpoint.
Worked Example: Zero Output Is the Best Feasible Choice
A hypothetical supplier may produce any quantity from zero to 5 units. Its revenue is \(TR(q)=2q\) dollars and its cost is \(TC(q)=q^2+5q+4\) dollars. Find the global profit maximum and explain why zero output does not mean zero profit.
Method. Compare endpoints and check the derivative over the whole interval. This avoids omitting the cost incurred at zero output or assuming that producing a positive quantity must improve profit.
Profit is:
Its derivative is \(\pi'(q)=-2q-3\). For every feasible \(q\ge0\), \(-2q-3<0\), so profit is strictly decreasing throughout the feasible interval. Consequently, its highest value occurs at the left endpoint. Evaluate both endpoints:
The global maximum is \(-4\) dollars at \(q=0\). In this model, \(TR(0)=0\) but \(TC(0)=4\), so \(\pi(0)=0-4=-4\), not zero. Since profit falls as output increases, no positive feasible output improves on the zero-output choice.
Check. The derivative is negative even at \(q=0\), where it equals \(-3\), and remains negative through \(q=5\), where it equals \(-13\). The endpoint comparison agrees with that direction of change.
Common Mistakes and Practical Checks
- Using the stationary condition as the final answer. The equation \(MR=MC\) identifies an interior candidate only. Check whether it is feasible and compare its profit with both endpoints.
- Ignoring zero output. The boundary \(q=0\) is feasible in this setup and must be evaluated, even when its profit is negative.
- Confusing a maximum with an interior maximum. A derivative condition does not apply at an endpoint. Profit can still be largest at a boundary because no further feasible movement in the improving direction is allowed.
- Reporting only one output when there is a tie. Compare the actual profit values. If two or more candidates share the largest value, all are global maximizers.
- Applying the differentiable-function rule to a kink. If the profit function is not differentiable at a feasible interior point, check that point directly as well as the endpoints and any differentiable stationary points.
A useful final check is to state both the maximizing output and its profit, with units, and to verify that the output is feasible. When output is restricted to whole units rather than any real quantity in an interval, evaluate every feasible integer output; a derivative-based stationary point need not be an available choice.
Check Your Understanding
For each problem, identify the feasible candidates and compare their profit values. Unless stated otherwise, output may be any real quantity in the specified interval.
- A hypothetical maker has \(\pi(q)=-q^2+8q-12\) for \(0\le q\le5\). Find the global maximizing output and its profit. (Answer: \(q=4\), profit \(4\).)
- A hypothetical service has \(\pi(q)=-q^2+6q-5\) for \(0\le q\le2\). Is its stationary output feasible, and what is the constrained maximum? (Answer: the stationary output is \(q=3\), which is infeasible; the maximum is \(3\) at \(q=2\).)
- A hypothetical plant has \(\pi(q)=-3q^2+12q-8\) for \(0\le q\le1\). Compare its endpoint profits and determine the maximum. (Answer: \(q=1\), profit \(1\).)
- A hypothetical firm has \(\pi(q)=q^3-6q^2+9q-1\) for \(0\le q\le4\). Find all feasible stationary points and all global maximizing outputs. (Answer: stationary points \(q=1,3\); global maxima at \(q=1\) and \(q=4\), each with profit \(3\).)
- A hypothetical producer can make only whole units \(q\in\{0,1,2,3,4\}\), with \(\pi(q)=-q^2+5q-4\). Find every maximizing feasible output and the maximum profit. (Answer: \(q=2\) and \(q=3\), each with profit \(2\).)