Reading the Marginal Curves
A graph of marginal revenue and marginal cost can show where increasing output stops adding to economic profit. The key is not simply to find a point where the curves meet. You must also check how the curves lie on either side of that point: an intersection is a profit maximum when marginal revenue is above marginal cost just before it and below marginal cost just after it.
This graphical method complements the earlier material on the marginal revenue equals marginal cost rule and on sufficient conditions for maximum profit. It also makes an important reading distinction visible: the output is read from the horizontal axis, the marginal values from the intersection’s vertical coordinate, and total revenue and total cost from their own curves at that output.
For a continuous-output model, the slope of the total-revenue curve at \(q\) is MR, and the slope of the total-cost curve is MC. Thus, if MR exceeds MC, a small increase in output adds more to revenue than to cost, so profit rises. If MR is less than MC, that increase adds more to cost than to revenue, so profit falls.
The usual interior-maximum check assumes that profit is differentiable near the candidate output and that small increases and decreases in output are feasible. Under those assumptions, an interior maximum must have \(\pi'(q)=0\), or equivalently \(MR(q)=MC(q)\). A crossing from MR above MC to MR below MC gives the local direction needed for a maximum. A tangency without a change in relative position does not, by itself, establish a maximum. For discrete output choices, compare the profit from the available quantities instead of treating the graph as continuous.
Why the Crossing Direction Matters
The direction test follows directly from the profit function. Subtracting the slope of total cost from the slope of total revenue gives the slope of profit. This explains both the intersection rule and the need to inspect the curves on either side.
Therefore, where MR is above MC, \(\pi'(q)>0\) and profit is increasing as output rises. Where MR is below MC, \(\pi'(q)<0\) and profit is decreasing as output rises. If the difference changes from positive to negative at an intersection, profit first rises and then falls; the intersection is a local maximum. This is a local conclusion. If the feasible output range has boundaries, those may also need comparison before calling an interior candidate the overall best choice.
A Reliable Graph-Reading Process
Read the axes and units before interpreting a crossing. The horizontal coordinate usually gives output, while the vertical coordinate gives dollars per unit (or another marginal measure). The intersection’s vertical coordinate is the value of MR and MC there; it is not total revenue, total cost, or profit.
Locate an intersection of the MR and MC curves and read its horizontal coordinate, \(q^*\).
Just to the left and right, compare which curve is higher. MR above MC means profit rises with output; MR below MC means profit falls with output.
At the intersection, read the shared vertical coordinate. This is both MR and MC at \(q^*\).
At the same horizontal coordinate, read TR and TC from their respective total curves. Then calculate profit as \(TR-TC\), if requested.
A graph may not give exact values. If it is a sketch, report an estimate and do not imply more precision than the scale supports. If equations or a detailed grid accompany the graph, use them to verify the visual reading. The marginal curves alone do not generally determine the levels of total revenue and total cost: their starting values or intercepts are also needed.
Worked Example: A Single Clear Crossing
A hypothetical equipment supplier’s graph plots output \(q\) in units on the horizontal axis and marginal dollars per unit on the vertical axis. The displayed curves are consistent with \(MR(q)=80-2q\) and \(MC(q)=20+q\), for \(0\le q\le 30\). The accompanying total curves are \(TR(q)=80q-q^2\) and \(TC(q)=100+20q+\frac{1}{2}q^2\). Find the profit-maximizing output, verify the direction around it, and read TR, TC, and profit there.
Method: locate the crossing and test its direction. Set the two marginal values equal:
Subtract 20 from both sides and add \(2q\) to both sides: \(60=3q\), so the curves intersect at \(q=20\). The vertical coordinate is \(MR(20)=80-2(20)=40\); also \(MC(20)=20+20=40\).
Check one output on each side. At \(q=19\), \(MR=80-38=42\) and \(MC=20+19=39\), so MR is above MC and profit is rising. At \(q=21\), \(MR=80-42=38\) and \(MC=20+21=41\), so MR is below MC and profit is falling. The crossing is therefore a local profit maximum.
Now read the total curves at \(q=20\): \(TR(20)=80(20)-20^2=1{,}600-400=1{,}200\). Also, \(TC(20)=100+20(20)+\frac{1}{2}(20^2)=100+400+200=700\). Thus profit is \(1{,}200-700=500\). As a check, at \(q=19\), TR is \(1{,}159\), TC is \(660.5\), and profit is \(498.5\); at \(q=21\), TR is \(1{,}239\), TC is \(740.5\), and profit is \(498.5\). Both nearby profits are lower than 500.
Common mistake avoided: The intersection’s vertical value, 40, is marginal revenue and marginal cost—not profit. Profit is found from the total curves and equals 500 at the selected output.
Worked Example: Using the Total Curves After Finding the Crossing
A hypothetical publisher’s graph shows \(MR(q)=100-4q\) and \(MC(q)=10+2q\), with output measured in batches. Its total curves are \(TR(q)=100q-2q^2\) and \(TC(q)=50+10q+q^2\), over \(0\le q\le 25\). Determine the maximizing output, confirm the direction of profit change, and calculate total revenue, total cost, and profit there.
Method: solve the marginal intersection, then read the total curves at that same output. Equating the marginal curves gives \(100-4q=10+2q\). Hence \(90=6q\) and \(q=15\). At that output, \(MR=100-4(15)=40\), and \(MC=10+2(15)=40\).
At \(q=14\), MR is \(100-56=44\) and MC is \(10+28=38\), so profit is rising as output approaches 15. At \(q=16\), MR is \(100-64=36\) and MC is \(10+32=42\), so profit is falling as output moves beyond 15. The intersection is a local maximum.
Evaluate total revenue: \(TR(15)=100(15)-2(15^2)=1{,}500-450=1{,}050\). Evaluate total cost: \(TC(15)=50+10(15)+15^2=50+150+225=425\). Profit is \(1{,}050-425=625\).
Check the neighboring integer outputs. At \(q=14\), TR is \(1{,}008\), TC is \(50+140+196=386\), and profit is \(622\). At \(q=16\), TR is \(1{,}088\), TC is \(50+160+256=466\), and profit is \(622\). Each is below the profit of 625 at \(q=15\).
Common mistake avoided: Do not use the MR or MC ordinate as the total amount. The shared marginal value is 40 per batch, while the total revenue and cost at \(q=15\) are 1,050 and 425.
Worked Example: Distinguishing Two Intersections
A hypothetical maker of custom fixtures can choose output only in the range \(3\le q\le 8\). Its marginal curves are \(MR(q)=8+10q-q^2\) and \(MC(q)=20+2q\). Its total curves are \(TR(q)=8q+5q^2-\frac{q^3}{3}\) and \(TC(q)=80+20q+q^2\). Use the graph and equations to identify the maximizing intersection within the feasible range and compare profit at the range’s endpoints.
Method: check the sign change at each intersection that matters. Equating MR and MC gives \(8+10q-q^2=20+2q\), or \(q^2-8q+12=0\). Factoring gives \((q-2)(q-6)=0\), so the curves intersect at \(q=2\) and \(q=6\). The first crossing is outside the feasible range, so it cannot be selected.
At \(q=5\), which is left of 6 but in the feasible range, \(MR=8+50-25=33\) and \(MC=20+10=30\); MR is above MC. At \(q=7\), which is right of 6, \(MR=8+70-49=29\) and \(MC=20+14=34\); MR is below MC. Profit rises toward \(q=6\) and falls after it, making 6 the interior local maximum.
At \(q=6\), \(TR=8(6)+5(36)-216/3=48+180-72=156\). Total cost is \(TC=80+20(6)+36=236\), so profit is \(156-236=-80\). At the lower endpoint \(q=3\), TR is \(24+45-9=60\), TC is \(80+60+9=149\), and profit is \(-89\). At the upper endpoint \(q=8\), TR is \(64+320-512/3=640/3\), TC is \(80+160+64=304\), and profit is \(640/3-304=-272/3\), approximately \(-90.67\). The interior output \(q=6\) gives higher profit than either endpoint.
Common mistake avoided: Treating every intersection as a maximum would wrongly include \(q=2\), which is not feasible here. Even for a feasible intersection, the curve positions on both sides—not equality alone—determine whether profit is rising into the point and falling afterward.
Common Graph-Reading Traps
- Stopping at equality. MR equals MC identifies a candidate, not automatically a maximum. Check that MR is above MC before the crossing and below it afterward.
- Reversing the direction test. If MR is above MC, an additional unit adds more revenue than cost, so profit rises. If MC is above MR, profit falls as output increases.
- Confusing marginal and total values. The vertical coordinate on an MR–MC graph is a marginal amount. Total revenue and total cost must be read from total curves or calculated from supplied total functions.
- Ignoring axis scale or units. A coordinate of 20 might mean 20 units, 20 batches, or another stated quantity. Preserve the graph’s units in the answer.
- Overstating precision. A rough drawn intersection may support an approximate output only. Use exact values only when the graph scale or accompanying equations justify them.
Check Your Understanding
For each graph, keep the marginal ordinate distinct from the total values. Answers are provided at the end of each question.
- A graph gives \(MR=64-2q\) and \(MC=16+q\). Find their intersection and state whether profit rises or falls as output passes through it. (Answer: \(q=16\), with \(MR=MC=32\); profit rises before 16 and falls after 16.)
- On a plotted graph, MR and MC cross at \(q=5\), with MR above MC just to the left and below MC just to the right. At \(q=5\), the total-revenue curve is 210 and the total-cost curve is 170. Identify the local maximum, calculate profit, and distinguish it from the marginal value if the intersection’s vertical coordinate is 30. (Answer: \(q=5\) is a local maximum; profit is \(210-170=40\); the marginal value is 30, not 40.)
- A graph is consistent with \(MR=120-3q\) and \(MC=12+q\). The total curves are \(TR=120q-1.5q^2\) and \(TC=90+12q+0.5q^2\). Find the intersection, test the direction using one output on each side, and calculate profit there. (Answer: \(q=27\), \(MR=MC=39\); at 26, MR \(=42\) exceeds MC \(=38\), and at 28, MR \(=36\) is below MC \(=40\); TR \(=2{,}146.5\), TC \(=778.5\), profit \(=1{,}368\).)
- A graph has crossings at \(q=4\) and \(q=9\). MR is below MC before 4, above MC between 4 and 9, and below MC after 9. Classify each crossing using the direction test. (Answer: \(q=4\) is a local minimum; \(q=9\) is a local maximum.)
- A graph’s feasible outputs include \(q=12\), where MR and MC intersect at a vertical coordinate of 48. At that output the total-revenue curve reads 960 and the total-cost curve reads 780. State the output, marginal values, and profit, and identify the tempting but incorrect interpretation of 48. (Answer: output \(q=12\); \(MR=MC=48\); profit \(=960-780=180\); 48 is not total profit.)