How Cost And Revenue Changes Alter Optimal Output

Why Separate Output from Profit?

A firm may face a change in the revenue it earns from each additional unit, a change in the cost of producing an additional unit, or both. These changes can alter the output that maximizes profit. But the direction and size of the output change do not, by themselves, tell us how much profit changes. To answer both questions, compare the marginal curves to identify the new best output, then evaluate the profit function at that output.

The analysis here uses a continuous-output model. Let $q\geq 0$ denote output, let $TR(q)$ be total revenue, and let $TC(q)$ be total cost. Profit is $\pi(q)=TR(q)-TC(q)$. Marginal revenue and marginal cost are the derivatives $MR(q)=TR'(q)$ and $MC(q)=TC'(q)$. We assume the relevant maximum is interior, so output is positive and the firm can vary it smoothly. Boundary choices must be checked separately if the calculated optimum is infeasible or lies outside the allowed output range.

Definition: An interior profit-maximizing output $q^*$ is a feasible output at which profit is highest among nearby feasible outputs. For differentiable revenue and cost functions, a necessary condition is $MR(q^*)=MC(q^*)$. A sufficient local condition is $\pi''(q^*)=MR'(q^*)-MC'(q^*)<0$. If profit is strictly concave over the relevant range, this interior maximum is unique and is the global maximum on that range.

How a Shift Changes the Marginal Condition

A change in marginal revenue or marginal cost changes the equation used to locate the best output. If marginal revenue shifts upward at every output while marginal cost is unchanged, the equality $MR=MC$ will generally occur at a larger output when marginal revenue slopes downward and marginal cost slopes upward. If marginal cost shifts upward while marginal revenue is unchanged, the equality will generally occur at a smaller output. These conclusions rely on the stated slope pattern and on an interior solution; if both curves shift, the net output change depends on their relative sizes.

The following simple linear model makes the comparison explicit. Suppose total revenue and total cost are $\displaystyle TR(q)=Aq-Bq^2$ and $\displaystyle TC(q)=Cq+Dq^2+F$, where $B>0$, $D>0$, and $F$ is a fixed cost. Then $MR(q)=A-2Bq$, $MC(q)=C+2Dq$, and profit is strictly concave because $\pi''(q)=-2(B+D)<0$. For an interior maximum, equating marginal revenue and marginal cost gives:

A-2Bq=C+2Dq \quad\Longrightarrow\quad q^*=\frac{A-C}{2(B+D)}

This formula applies when its solution is positive and feasible. It shows the separate roles of the shifts: raising $A$, the intercept of marginal revenue, raises $q^*$; raising $C$, the intercept of marginal cost, lowers $q^*$; and increasing $B$ or $D$ makes the relevant marginal curves steeper, reducing the quantity response to a given intercept difference. The fixed cost $F$ does not appear in the formula because its derivative with respect to output is zero.

Why the Marginal Rule Works

The marginal condition follows from the way a small output change affects profit. Since $\pi(q)=TR(q)-TC(q)$, the derivative of profit is $\pi'(q)=MR(q)-MC(q)$. If $MR>MC$, an additional small unit adds more to revenue than to cost, so a small increase in output raises profit. If $MR<MC$, that increase lowers profit, and reducing output raises profit. At an interior maximum, neither a small increase nor a small decrease should improve profit, so the first derivative is zero and $MR=MC$.

That equality alone is not enough to identify a maximum: it could describe a minimum or another stationary point. In the linear model above, however, the second derivative is $-2(B+D)$, which is negative because $B,D>0$. Profit therefore curves downward throughout the range, and the stationary point is the unique maximum. This is why the model's assumptions matter when using the marginal equality.

Write the changed functions.
Keep the original and revised revenue and cost functions distinct so that each shift is visible.

Calculate the marginal functions.
Differentiate total revenue and total cost to obtain $MR$ and $MC$ in each case.

Find and check the output.
Solve $MR=MC$, then verify that the solution is feasible and that the second-order condition supports a maximum.

Evaluate profit separately.
Substitute each optimal output into its corresponding $TR-TC$. If useful, also calculate revised profit at the old output to isolate the effect of re-optimizing.

Worked Comparisons

The examples use hypothetical firms and amounts in dollars per operating period. Each revenue and cost function is assumed to apply over a feasible range containing the calculated interior optimum. The calculations compare both output and profit rather than treating one as a substitute for the other.

Worked Example: An Increase in Marginal Cost

A hypothetical firm initially has $TR_0(q)=80q-q^2$ and $TC_0(q)=8q+q^2$. A supplier-cost change revises total cost to $TC_1(q)=16q+q^2$, leaving revenue unchanged. Find the old and new optimal outputs and maximum profits. Also find revised profit if the firm mistakenly kept producing its old output.

Method. Differentiate the functions, solve $MR=MC$ in each case, verify concavity, and then calculate profit at the relevant quantities. This avoids mistaking the cost change at the old output for the change in maximum profit.

Initially, $MR(q)=80-2q$ and $MC_0(q)=8+2q$. Thus $80-2q=8+2q$, so $72=4q$ and $q_0^*=18$. Profit is $\pi_0(q)=72q-2q^2$, and $\pi_0(18)=72(18)-2(18^2)=1{,}296-648=\$648$. Its second derivative is $-4$, so this is a maximum.

After the change, $MC_1(q)=16+2q$. Solving $80-2q=16+2q$ gives $64=4q$, hence $q_1^*=16$. The revised profit function is $\pi_1(q)=64q-2q^2$. At the new optimum, $\pi_1(16)=64(16)-2(16^2)=1{,}024-512=\$512$, again a maximum because the second derivative is $-4$.

If output stayed at 18, revised profit would be $\pi_1(18)=64(18)-2(18^2)=1{,}152-648=\$504$. The cost increase at that fixed output is $TC_1(18)-TC_0(18)=(16(18)+18^2)-(8(18)+18^2)=612-468=\$144$, reducing profit from $\$648$ to $\$504$. Re-optimizing raises profit by $\$8$ relative to that fixed-output result, to $\$512$. Overall, output falls by 2 units and maximum profit falls by $\$136$.

Worked Example: An Increase in Marginal Revenue

A hypothetical firm has $TR_0(q)=60q-q^2$ and $TC(q)=12q+q^2$. A change in its sales environment revises revenue to $TR_1(q)=72q-q^2$, with cost unchanged. Determine the old and new optimal outputs and maximum profits, and compare revised profit at the old output with revised maximum profit.

Method. Find each stationary output from the corresponding marginal equality, then evaluate both profit functions. The key check is to use the revised revenue function when calculating revised profit.

Initially, $MR_0(q)=60-2q$ and $MC(q)=12+2q$. Equating them gives $60-2q=12+2q$, so $48=4q$ and $q_0^*=12$. Initial profit is $\pi_0(q)=48q-2q^2$, giving $\pi_0(12)=576-288=\$288$. The second derivative is $-4$, confirming a maximum.

After the revenue change, $MR_1(q)=72-2q$. The new equality is $72-2q=12+2q$, so $60=4q$ and $q_1^*=15$. Revised profit is $\pi_1(q)=60q-2q^2$, and $\pi_1(15)=900-450=\$450$. Its second derivative is also $-4$, so this is the maximum.

At the old output of 12, revised profit would be $\pi_1(12)=60(12)-2(12^2)=720-288=\$432$. The revenue improvement at that quantity is $TR_1(12)-TR_0(12)=12(12)=\$144$, so fixed-output profit rises from $\$288$ to $\$432$. Increasing output from 12 to 15 raises revised profit by a further $\$18$, from $\$432$ to $\$450$. Thus output rises by 3 units and maximum profit rises by $\$162$.

Worked Example: Revenue and Cost Both Change

A hypothetical producer initially has $TR_0(q)=100q-q^2$ and $TC_0(q)=20q+q^2$. A redesign raises revenue to $TR_1(q)=104q-q^2$, while a production change raises cost to $TC_1(q)=20q+2q^2$. Find both optimal outputs and maximum profits. Explain why the direction of the output change cannot be inferred from the revenue change alone.

Method. Use the new marginal functions together; do not assume that one curve's shift determines the outcome when the other also changes.

Initially, $MR_0(q)=100-2q$ and $MC_0(q)=20+2q$. Equating them gives $80=4q$, so $q_0^*=20$. Initial profit is $\pi_0(q)=80q-2q^2$, and $\pi_0(20)=1{,}600-800=\$800$. The second derivative is $-4$, confirming a maximum.

After both changes, $MR_1(q)=104-2q$ and $MC_1(q)=20+4q$. The new equality is $104-2q=20+4q$, which gives $84=6q$, or $q_1^*=14$. Revised profit is $\pi_1(q)=84q-3q^2$. Therefore, $\pi_1(14)=84(14)-3(14^2)=1{,}176-588=\$588$. Since $\pi_1''(q)=-6$, this is a maximum.

At the old output of 20, revised profit is $\pi_1(20)=84(20)-3(20^2)=1{,}680-1{,}200=\$480$. Re-optimizing raises revised profit by $\$108$, from $\$480$ to $\$588$, but it does not restore the original $\$800$. Output falls by 6 units and maximum profit falls by $\$212$. Although the revenue change raises marginal revenue, the marginal cost curve also becomes steeper; the combined changes produce a lower intersection quantity.

Output Effects and Profit Effects Are Different

A change in the best output is a response to the new $MR=MC$ intersection. A change in maximum profit compares the height of the profit function at the old maximum with its height at the new maximum. Those are separate calculations. Even when a firm increases output after a revenue improvement, the gain in profit depends on the revised revenue and cost functions over the whole production level, not just on the number of additional units.

It is also useful to distinguish a change in marginal cost from a fixed-cost change. If total cost becomes $TC_1(q)=TC_0(q)+K$, where $K$ is a constant, then $MC_1(q)=MC_0(q)$. The interior marginal condition and its output are unchanged, while profit at every output falls by $K$. This statement concerns the unconstrained interior condition; it does not determine whether production remains feasible or optimal when other constraints apply.

Key takeaway: Recalculate the marginal condition after each change to find the new output, then substitute that output into the corresponding total revenue and total cost functions. Never infer the change in maximum profit from the output change alone.

Common Mistake: Mixing the Old and New Functions

A frequent error is to find a new output correctly but evaluate its profit with the old cost function, or to calculate the change in profit at the old output and call it the change in maximum profit. Both calculations answer a different question.

For example, in the first worked example, revised profit at the old output is $TR_0(18)-TC_1(18)=1{,}440-612=\$828$ only if one mistakenly uses the wrong revenue value; the correct revenue at 18 is $80(18)-18^2=1{,}116$, so revised profit at 18 is $1{,}116-612=\$504$. The new maximum is $\$512$ at output 16. The incorrect calculation mixes a miscomputed revenue with the revised cost; the correction is to evaluate both revised functions at the same quantity.

Another trap is to report only the new quantity. A complete comparison identifies the old and new quantities, calculates profit under the appropriate functions at each optimum, and states whether the profit change is positive or negative. If the question asks about holding output fixed, calculate that counterfactual separately.

Check Your Understanding

For each problem, assume the stated functions apply over a feasible range containing the interior maximum. Calculate output and profit using the functions for the relevant case.

A firm has $TR(q)=70q-q^2$ and initially $TC_0(q)=10q+q^2$. Cost changes to $TC_1(q)=18q+q^2$. Find both optimal outputs and maximum profits. (Answer: output falls from 15 to 13 units; maximum profit falls from $\$450$ to $\$338$.)
A firm has $TR_0(q)=50q-q^2$, revised revenue $TR_1(q)=62q-q^2$, and unchanged $TC(q)=6q+q^2$. Find both optimal outputs and maximum profits. (Answer: output rises from 11 to 14 units; maximum profit rises from $\$242$ to $\$392$.)
A firm has $TR(q)=90q-q^2$, with $TC_0(q)=10q+q^2$ and revised $TC_1(q)=10q+1.5q^2$. Find both optimal outputs and maximum profits. (Answer: output falls from 20 to 16 units; maximum profit falls from $\$800$ to $\$640$.)
A firm initially has $TR_0(q)=80q-q^2$ and $TC_0(q)=8q+q^2$. The revised functions are $TR_1(q)=88q-q^2$ and $TC_1(q)=12q+1.5q^2$. Find both optimal outputs and maximum profits. (Answer: output falls from 18 to 15.2 units; maximum profit falls from $\$648$ to $\$577.60$.)
A firm has $TR(q)=40q-q^2$ and $TC_0(q)=8q+q^2$. A fixed cost of $\$30$ is added, so $TC_1(q)=8q+q^2+30$. Find the optimal output and profit before and after the change. (Answer: output remains 8 units; profit falls from $\$128$ to $\$98$.)

What You'll Learn