Corrected exercises: marginal analysis (201-103)

$C'(q)=0.6q+25$, so $C'(60)=36+25=61$ dollars. The fixed cost $600$ disappears on differentiation: it does not weigh on the next unit.

$C'(q)=0.03q^{2}-1.2q+40$, so $C'(30)=0.03\times 900-1.2\times 30+40=27-36+40=31$ dollars.

$C(q)=800+60q^{\frac{1}{2}}+4q$, so $C'(q)=30q^{-\frac{1}{2}}+4=\frac{30}{\sqrt{q}}+4$. At $q=36$: $C'(36)=\frac{30}{6}+4=9$ dollars. Pitfall: differentiating $60\sqrt{q}$ as $\frac{60}{\sqrt{q}}$, forgetting the $\frac{1}{2}$ of the exponent.

$C'(q)=12+0.4\times\frac{3}{2}q^{\frac{1}{2}}=12+0.6\sqrt{q}$. At $q=64$: $C'(64)=12+0.6\times 8=16.8$ dollars.

$R(q)=q(120-0.4q)=120q-0.4q^{2}$, so $R(100)=12{,}000-4{,}000=8{,}000$ dollars. $R'(q)=120-0.8q$, so $R'(100)=40$ dollars. Marginal revenue falls twice as fast as the price: at $q=100$ the price is $80$ and the marginal revenue $40$.

The price is $p(100)=60-2\times 10=40$ dollars. $R(q)=q(60-2q^{\frac{1}{2}})=60q-2q^{\frac{3}{2}}$, so $R'(q)=60-3q^{\frac{1}{2}}=60-3\sqrt{q}$ and $R'(100)=60-30=30$ dollars. Pitfall: taking the price $40$ for the marginal revenue, or differentiating $p$ instead of $R$.

$\overline{C}(q)=0.1q+12+\frac{2{,}250}{q}$, so $\overline{C}(50)=5+12+45=62$ dollars. $\overline{C}'(q)=0.1-\frac{2{,}250}{q^{2}}$, so $\overline{C}'(50)=0.1-\frac{2{,}250}{2{,}500}=0.1-0.9=-0.8$. Negative: at $50$ units, the average cost is still falling. Pitfall: differentiating $\frac{2{,}250}{q}$ as $\frac{2{,}250}{1}$; it is $2{,}250q^{-1}$, whose derivative is $-2{,}250q^{-2}$.

$\overline{C}(q)=0.4q+16+\frac{1{,}000}{q}$ and $\overline{C}'(q)=0.4-\frac{1{,}000}{q^{2}}=0$ gives $q^{2}=2{,}500$, that is $q=50$ for $q>0$. The minimum average cost is $\overline{C}(50)=20+16+20=56$ dollars. Check: $C'(50)=0.8\times 50+16=56$ as well, since marginal cost crosses average cost at its minimum.

$R(q)=150q-0.5q^{2}$, so $R'(q)=150-q$; $C'(q)=0.5q+30$. At $q=60$: $R'(60)=90$ and $C'(60)=60$, so $P'(60)=90-60=30$ dollars. Positive: producing one more unit still increases the profit.

$R(q)=200q-0.6q^{2}$, so $R'(q)=200-1.2q$; $C'(q)=0.8q+40$. $200-1.2q=0.8q+40$ gives $2q=160$, that is $q=80$. The selling price is then $p(80)=200-48=152$ dollars. Pitfall: giving $R'(80)=104$ as the price; the price is read off the DEMAND.

The constant term 800 is the FIXED COST: it is paid even if nothing is produced, such as rent, insurance and depreciation. The term $20q$ is a PROPORTIONAL variable cost: each unit costs 20 dollars in materials, whatever the quantity. The term $0.5q^{2}$ is a variable cost that accelerates: it expresses that as production grows, each additional unit becomes more expensive, through overtime, saturated machines or the use of costlier suppliers.

$C'(q)=q+20$. At $q=40$: $C'(40)=40+20=60$ dollars. The marginal cost is therefore 60 dollars at a production level of 40 units, and it is also the slope of the tangent to the cost curve at that point, as the figure shows.

$C(41)=0.5\times 1{,}681+20\times 41+800=840.5+820+800=2{,}460.5$. $C(40)=0.5\times 1{,}600+800+800=800+800+800=2{,}400$. The exact cost of the 41st unit is therefore $2{,}460.5-2{,}400=60.50$ dollars, against the 60 dollars announced by the derivative. The gap, 50 cents, comes from the derivative being the slope of the tangent and not that of the chord: it approximates the cost of the next unit all the better when the function is only slightly curved and $q$ is large.

Average cost: $\overline{C}(q)=\dfrac{C(q)}{q}=0.5q+20+\dfrac{800}{q}$. At $q=40$: $\overline{C}(40)=20+20+20=60$ dollars per unit. Note that the average cost and the marginal cost are both 60 at $q=40$: this is no coincidence, and exercise 4 will give the reason.

The marginal cost $q+20$ is increasing because the quadratic term of the total cost has a POSITIVE coefficient. Economically, this is the law of DIMINISHING RETURNS: beyond a certain point, adding resources to a fixed production capacity yields less and less output per dollar spent. A factory already running at full capacity has to pay overtime, rent extra space or accept more scrap. Without this increase, the optimal production would be infinite, which is enough to show that the assumption is necessary to the model.

$R(q)=p\times q=\left(100-0.5q\right)q=100q-0.5q^{2}$. At $q=40$: $R(40)=4{,}000-0.5\times 1{,}600=4{,}000-800=3{,}200$ dollars, with a unit price of $100-20=80$ dollars.

$R'(q)=100-q$. At $q=40$: $R'(40)=100-40=60$ dollars. Selling a 41st unit would therefore bring in about 60 dollars more, not 80 dollars, even though 80 dollars is what the customer would pay.

Because selling one more unit forces the price down on ALL the units sold, not only on the last one. The price falls by 0.5 dollar per additional unit: you gain the price of the new unit, about $100-0.5q$, and you lose $0.5q$ on the $q$ previous units. In total $\left(100-0.5q\right)-0.5q=100-q$, which is exactly the marginal revenue. The slope of the marginal revenue, $-1$, is therefore double that of the demand, $-0.5$, and this relation holds for every linear demand.

The revenue is maximal when $R'(q)=0$, so $100-q=0$ and $q=100$. The revenue is then $R(100)=10{,}000-5{,}000=5{,}000$ dollars, at a price of $100-50=50$ dollars per unit. We check that it is indeed a maximum: $R''(q)=-1<0$, the parabola opens downward.

Beyond $q=100$, the marginal revenue becomes NEGATIVE: each additional unit loses revenue, because the price cut imposed on the whole stock exceeds what the new unit brings in. Producing more would therefore bring in less, and cost more. No rational firm places itself in this zone, and the next exercise will show that the maximum profit lies well before, at $q=40$.

$P(q)=\left(100q-0.5q^{2}\right)-\left(0.5q^{2}+20q+800\right)=100q-0.5q^{2}-0.5q^{2}-20q-800=80q-q^{2}-800$.

$P'(q)=80-2q$, which vanishes at $q=40$. Check by the second derivative: $P''(q)=-2<0$, so the function is concave and the critical point is indeed a MAXIMUM. It can also be seen from the sign of $P'$: positive before 40, negative after, so the function rises and then falls.

Marginal revenue at 40: $R'(40)=100-40=60$. Marginal cost at 40: $C'(40)=40+20=60$. Both equal 60 dollars: the equality holds. The interpretation is direct: at 40 units, the 41st would bring in 60 dollars and cost 60 dollars, so it would add nothing.

The profit is $P=R-C$, so $P'=R'-C'$. A necessary condition for a maximum is $P'(q)=0$, that is $R'(q)-C'(q)=0$, so $R'(q)=C'(q)$: marginal revenue equals marginal cost. The sufficient condition comes on top: $P''(q)=R''(q)-C''(q)<0$, in other words marginal cost must grow faster than marginal revenue at the point considered. Here $R''=-1$ and $C''=1$, so $P''=-2<0$, and the condition is met. This rule, known as “marginal revenue equals marginal cost”, is the most quoted result in all of microeconomics, and it is nothing other than a derivative set to zero.

$P(40)=80\times 40-1{,}600-800=3{,}200-1{,}600-800=800$ dollars. Selling price: $p=100-0.5\times 40=80$ dollars. To remember: the quantity that maximizes PROFIT, 40, is well below the one that maximizes REVENUE, 100. Maximizing sales and maximizing profit are two different goals, and it is the cost that separates them.

$\overline{C}'(q)=0.5-\dfrac{800}{q^{2}}$. Set it to zero: $0.5=\dfrac{800}{q^{2}}$, so $q^{2}=1{,}600$ and $q=40$, the negative root being rejected. Check: $\overline{C}''(q)=\dfrac{1{,}600}{q^{3}}>0$ for $q>0$, the function is convex and the critical point is a MINIMUM.

$\overline{C}(40)=0.5\times 40+20+\dfrac{800}{40}=20+20+20=60$ dollars per unit. $C'(40)=40+20=60$ dollars. The two coincide, as the crossing of the two curves on the figure shows.

Write $\overline{C}(q)=\dfrac{C(q)}{q}$ and differentiate by the quotient rule: $\overline{C}'(q)=\dfrac{C'(q)q-C(q)}{q^{2}}$. Setting this derivative to zero amounts to setting the numerator to zero, so $C'(q)q=C(q)$, that is $C'(q)=\dfrac{C(q)}{q}=\overline{C}(q)$. At the minimum of the average cost, the marginal cost therefore equals the average cost, and the proof assumes nothing about the form of $C$: it holds for every differentiable cost function.

The image is that of a grade average. If the mark on the next test, the marginal, is BELOW the current average, the average goes down. If it is above, the average goes up. The average is therefore lowest precisely when the new mark equals it. This is exactly what the graph says: as long as the marginal cost is below the average cost, the latter decreases; once it overtakes it, the average cost rises again; the crossing is the minimum.

That the firm is profitable at the optimal point: it sells for 80 dollars what costs it 60 dollars on average, a margin of 20 dollars per unit, which gives back $20\times 40=800$ dollars of profit. Beware, though: the fact that the average cost is minimal at $q=40$ does not make the profit maximal there; the two coincide here by construction of the example. In general, the quantity of minimum average cost and that of maximum profit are DIFFERENT, and it is the second that governs the decision.

The linear approximation reads $\Delta C\approx C'(q)\Delta q$. Here $C'(40)=60$ and $\Delta q=3$, so $\Delta C\approx 60\times 3=180$ dollars.

Exactly: $C(43)=0.5\times 1{,}849+20\times 43+800=924.5+860+800=2{,}584.5$, and $C(40)=2{,}400$, so $\Delta C=184.50$ dollars. The error is $184.50-180=4.50$ dollars, which is 2.4% of the change. The approximation always UNDERESTIMATES here, because the function is convex and the tangent lies below the curve.

Geometrically, $C'(q)\Delta q$ is the height gained by following the TANGENT, and $\Delta C$ the height gained by following the CURVE. On a convex function, the curve pulls away from the tangent upward, and the gap grows like the square of the displacement. This is why the approximation is excellent for one unit and deteriorates quickly when you extrapolate over three, ten or twenty units.

Expand: $C(q+1)-C(q)=0.5\left(q+1\right)^{2}+20\left(q+1\right)+800-\left(0.5q^{2}+20q+800\right)=0.5\left(2q+1\right)+20=q+0.5+20$. Now $C'(q)=q+20$. The gap is therefore $\left(q+20.5\right)-\left(q+20\right)=0.5$ dollar, whatever the quantity. This result comes from the quadratic coefficient: for $C(q)=aq^{2}+bq+c$, the one-unit gap is always exactly $a$, here 0.5.

On a LINEAR cost function, $C(q)=bq+c$, the tangent is the curve itself: the approximation is EXACT and the error is zero for every $\Delta q$. On a very curved function, with a large second derivative, the error becomes significant from the first unit: this is the case of costs that explode near a factory's maximum capacity. The practical rule is to rely on the marginal cost only for changes that are small compared with the scale on which the curvature shows up.

$-q^{2}+80q-800=0$, that is $q^{2}-80q+800=0$. Discriminant: $80^{2}-4\times 800=6{,}400-3{,}200=3{,}200$, so $\sqrt{3{,}200}=40\sqrt{2}\approx 56.5685$. Roots: $q=\dfrac{80\pm 56.5685}{2}$, that is $q_{1}\approx 11.72$ and $q_{2}\approx 68.28$.

The parabola $P(q)=-q^{2}+80q-800$ opens downward, so it is POSITIVE between its roots: the firm is profitable for $q$ between 11.72 and 68.28 units, in practice from 12 to 68 units. Outside this interval, it loses money.

At 10 units: $P(10)=800-100-800=-100$ dollars, the firm loses 100 dollars, because the output is not enough to cover the 800 dollars of fixed cost. At 75 units: $P(75)=6{,}000-5{,}625-800=-425$ dollars, it loses 425 dollars, this time because the marginal cost overtook the marginal revenue long ago: producing too much costs as dearly as producing too little.

The profit becomes $P(q)=80q-q^{2}-1{,}200$. Roots of $q^{2}-80q+1{,}200=0$: discriminant $6{,}400-4{,}800=1{,}600$, root 40, so $q=\dfrac{80\pm 40}{2}$, that is 20 and 60. The interval of profitability NARROWS, from the range 11.7 to 68.3 to the range 20 to 60. The maximum profit, for its part, is still reached at $q=40$, since the fixed cost does not change the derivative, but it drops to $3{,}200-1{,}600-1{,}200=400$ dollars. A fixed cost moves the profit down without moving the optimum: this is a very general property, and often counterintuitive.

The maximum profit is $P(40)=3{,}200-1{,}600-F=1{,}600-F$ where $F$ is the fixed cost. It becomes negative as soon as $F>1{,}600$ dollars. Beyond this threshold, the firm loses money whatever its output, and the rational long-run decision is to shut down. In the short run, the reasoning differs: as long as the price covers the average variable cost, producing limits the losses, since the fixed cost is paid anyway.

$Q'(L)=-L^{2}+24L$. This is a parabola opening downward, maximal when its derivative vanishes: $Q''(L)=-2L+24=0$, so $L=12$. The marginal product there is $Q'(12)=-144+288=144$ units per additional employee.

$Q''(L)=-2L+24$ vanishes at $L=12$ and CHANGES SIGN: positive before, negative after. The concavity of $Q$ therefore changes at this point, which is the definition of an inflection point. Its value: $Q(12)=-\dfrac{1{,}728}{3}+12\times 144=-576+1{,}728=1{,}152$ units.

BEFORE 12 employees, the marginal product INCREASES: each new employee brings in more than the previous one, because specialization and the division of labour become possible. This is the phase of increasing returns. AFTER 12 employees, the marginal product DECREASES: each new employee still brings in something, but less than the previous one, because the equipment and the space are fixed. This is the law of diminishing returns, and the inflection point marks exactly where it begins.

Total output is maximal when $Q'(L)=0$, so $-L^{2}+24L=0$, that is $L\left(24-L\right)=0$ and $L=24$, the solution $L=0$ being a minimum. Output: $Q(24)=-\dfrac{13{,}824}{3}+12\times 576=-4{,}608+6{,}912=2{,}304$ units. Beyond 24 employees, the marginal product becomes negative: adding people makes the total output FALL, which describes a saturated assembly line rather well.