Corrected exercises: elasticity and economic optimization (201-103)

$\frac{dq}{dp}=-3$ and, at $p=10$, $q=90$. So $E=-3\times\frac{10}{90}=-\frac{1}{3}\approx-0.33$: $|E|<1$, so the demand is inelastic. Pitfall: answering $-3$, the slope, instead of the elasticity.

$E=\frac{-6p}{480-6p}=\frac{-p}{80-p}$. Solve $\frac{p}{80-p}=1$: $p=80-p$, so $p=40$. A linear demand has unit elasticity at the MIDPOINT of its price segment, here between $0$ and $80$.

$q=600p^{-1}$, so $\frac{dq}{dp}=-\frac{600}{p^{2}}$, and $E=-\frac{600}{p^{2}}\cdot\frac{p}{600/p}=-1$, whatever $p$ is. At $p=12$: $E=-1$, unit elasticity. The spending $pq=600$ does not depend on the price, and that is exactly what $E=-1$ says.

$q=1{,}800p^{-2}$, so $\frac{dq}{dp}=-3{,}600p^{-3}$, and $E=-3{,}600p^{-3}\cdot\frac{p}{1{,}800p^{-2}}=-2$, whatever $p$ is. At $p=6$: $q=50$, $\frac{dq}{dp}=-\frac{3{,}600}{216}\approx-16.67$ and $E=-16.67\times\frac{6}{50}=-2$: the demand is elastic. Rule: for $q=kp^{-n}$, the elasticity equals $-n$ everywhere.

$\frac{dq}{dp}=-4p$. At $p=8$: $q=300-128=172$ and $\frac{dq}{dp}=-32$, so $E=-32\times\frac{8}{172}=-\frac{256}{172}\approx-1.49$. Pitfall: squaring $2\times 8$ when evaluating $q=300-2\times 8^{2}$, which gives $q=44$.

$\frac{dq}{dp}=\frac{-1}{2\sqrt{100-p}}$, so $E=\frac{-1}{2\sqrt{100-p}}\cdot\frac{p}{\sqrt{100-p}}=\frac{-p}{2(100-p)}$. At $p=64$: $E=-\frac{64}{72}=-\frac{8}{9}\approx-0.89$, inelastic demand. Pitfall: forgetting the minus sign, the derivative of $100-p$, which would give an increasing demand.

The demand is given backwards: $\frac{dp}{dq}=-0.25$, so $\frac{dq}{dp}=\frac{1}{-0.25}=-4$. At $q=120$: $p=80-30=50$, and $E=-4\times\frac{50}{120}=-\frac{5}{3}\approx-1.67$. Pitfall: taking $-0.25$ as $\frac{dq}{dp}$, which gives $E\approx-0.10$.

$\frac{dq}{dR}=-0.0005$ and, at $R=20{,}000$, $q=40-10=30$. So $E_{R}=-0.0005\times\frac{20{,}000}{30}=-\frac{10}{30}\approx-0.33$. A NEGATIVE income elasticity signals an inferior good: people buy less of it when income rises.

$\frac{dq_{A}}{dp_{B}}=-2$ and, at $p_{B}=30$, $q_{A}=240$. So $E_{c}=-2\times\frac{30}{240}=-0.25$. A NEGATIVE cross-price elasticity signals complementary goods: when $B$ becomes more expensive, people buy less $B$ and therefore less $A$.

At $p=10$, $q=300$; at $p=12$, $q=256$. So $\Delta q=-44$, $\overline{q}=278$, $\Delta p=2$, $\overline{p}=11$, and $E_{arc}=\frac{-44/278}{2/11}=-\frac{44\times 11}{278\times 2}=-\frac{484}{556}\approx-0.87$. Pitfall: dividing by the STARTING values, $300$ and $10$, which gives $-0.73$; the arc is measured against the AVERAGES so that it gives the same number in both directions.

$\dfrac{dq}{dp}=-2$, so $E=-2\times \dfrac{p}{200-2p}=\dfrac{-2p}{200-2p}=\dfrac{-p}{100-p}$. Dividing through by 2 is convenient and holds for every $p<100$.

$E(40)=\dfrac{-40}{60}\approx -0.667$; $E(50)=\dfrac{-50}{50}=-1$; $E(60)=\dfrac{-60}{40}=-1.5$. So the demand is INELASTIC at 40 dollars, UNIT elastic at 50 dollars, and ELASTIC at 60 dollars.

It is negative because the demand is decreasing: a price increase makes the quantity fall, so $\dfrac{dq}{dp}<0$. Many authors report the absolute value to avoid confusion, and when they do they must say so explicitly. It has no unit because it is a ratio of two RELATIVE CHANGES: the units of $q$ cancel each other, and so do those of $p$. That is what lets us compare the elasticity of the demand for gasoline with that of the demand for movie tickets, which a slope could never do.

No, the elasticity VARIES all along a demand that is nonetheless linear: it equals 0 at $p=0$, $-1$ at the midpoint, and tends to infinity as $p$ approaches 100. The reason is in the formula: the slope is constant, but the factor $\dfrac{p}{q}$ is not, and it blows up as $q$ becomes small. This distinction between slope and elasticity is the most misunderstood point of the chapter: a line does not have one elasticity, it has one per point.

At $p=40$, $q=120$; at $p=44$, $q=200-88=112$. So $\Delta q=-8$, $\Delta p=4$, $\overline{q}=\dfrac{120+112}{2}=116$ and $\overline{p}=\dfrac{40+44}{2}=42$. Then $E_{arc}=\dfrac{-8/116}{4/42}=\dfrac{-0.068966}{0.095238}\approx -0.724$. This is close to $E(40)=-0.667$ without being identical: the arc elasticity measures a FINITE change and is referred to the midpoint of the interval, while the derivative measures an instantaneous change at a point. The two get closer as the interval shrinks, and the arc is what you use when you only have two observations.

$R(p)=pq=p\left(200-2p\right)=200p-2p^{2}$. Then $R'(p)=200-4p$, which vanishes at $p=50$. Since $R''(p)=-4<0$, this is a maximum, and $R(50)=10{,}000-5{,}000=5{,}000$ dollars.

At $p=50$: $q=200-100=100$ and $E=\dfrac{-50}{100-50}=-1$. So the elasticity equals exactly $-1$ at the peak of revenue, which is the central result of the chapter.

$R=pq$ where $q$ depends on $p$. By the product rule: $\dfrac{dR}{dp}=q+p\dfrac{dq}{dp}$. Factor out $q$: $\dfrac{dR}{dp}=q\left(1+\dfrac{p}{q}\dfrac{dq}{dp}\right)=q\left(1+E\right)$, since $E=\dfrac{dq}{dp}\dfrac{p}{q}$ by definition. The proof takes two lines and assumes nothing about the form of the demand.

Since $q>0$, the sign of $\dfrac{dR}{dp}$ is that of $1+E$. If the demand is INELASTIC, $-1<E<0$, so $1+E>0$: raising the price INCREASES revenue, because the quantity lost is proportionally smaller than the price increase. If it is ELASTIC, $E<-1$, so $1+E<0$: raising the price LOWERS revenue. And if $E=-1$, revenue is stationary, which is the maximum. This rule explains why the price of tobacco gets raised and not that of airline tickets.

Yes, it should be raised. At $p=40$, $E\approx -0.667$, the demand is inelastic, so $1+E>0$ and revenue grows with the price. In figures: $R(40)=40\times 120=4{,}800$ dollars and $R(50)=50\times 100=5{,}000$ dollars, which is 200 dollars more. Note that the quantity sold FALLS, from 120 to 100 units: you sell less and take in more. Be careful, though: maximizing revenue is not maximizing profit, since cost does not appear anywhere here.

Income elasticity is $E_{R}=\dfrac{dq}{dR}\times \dfrac{R}{q}$: the relative change in the quantity demanded for a relative change in the consumer's income. Here $\dfrac{dq}{dR}=0.001$ and, at $R=40{,}000$, $q=50+40=90$. So $E_{R}=0.001\times \dfrac{40{,}000}{90}=\dfrac{40}{90}\approx 0.444$.

If $E_{R}<0$, the good is INFERIOR: people buy less of it when income rises, as with cheap pasta or, in some contexts, public transit. If $0<E_{R}<1$, the good is NORMAL and a necessity: consumption rises, but more slowly than income, which is the case here with 0.444. If $E_{R}>1$, the good is a LUXURY: its consumption grows faster than income, as with travel or restaurants. Engel's law, which says that the share of the budget spent on food decreases as income rises, is nothing other than an income elasticity below 1.

Cross-price elasticity is $E_{c}=\dfrac{dq_{A}}{dp_{B}}\times \dfrac{p_{B}}{q_{A}}$: it measures the sensitivity of the demand for one good to the price of ANOTHER good. Here $\dfrac{dq_{A}}{dp_{B}}=0.8$ and, at $p_{B}=50$, $q_{A}=120+40=160$. So $E_{c}=0.8\times \dfrac{50}{160}=\dfrac{40}{160}=0.25$.

A POSITIVE sign, as here, signals SUBSTITUTE goods: when the price of $B$ rises, people switch to $A$, whose demand increases; this is the case with butter and margarine, or with train and plane. A NEGATIVE sign signals COMPLEMENTARY goods: when the price of $B$ rises, people buy less of it and also less of $A$, as with printers and ink cartridges. A value close to zero signals unrelated goods. The value of 0.25 obtained here indicates a real but weak substitutability.

Because elasticity measures a capacity to ADJUST, and adjusting takes time. In the short run, a rise in the price of gasoline changes almost nothing: you still have to get to work, the vehicle is already bought, the trip is the same. In the long run, people change vehicles, move house, change their travel habits, and the drop in consumption becomes much larger. The general rule holds for almost every good: elasticity is larger in absolute value in the long run than in the short run. That is why a gasoline tax brings in a lot of revenue the first year and less and less afterwards.

Let $x$ be the side parallel to the wall and $y$ each of the two perpendicular sides. The fencing covers three sides: $x+2y=240$. The area to maximize is $A=xy$.

From the constraint, $x=240-2y$, so $A(y)=\left(240-2y\right)y=240y-2y^{2}$, for $0<y<120$. Then $A'(y)=240-4y$, which vanishes at $y=60$. Since $A''(y)=-4<0$, this is indeed a maximum.

$y=60$ metres and $x=240-120=120$ metres. Area: $A=120\times 60=7{,}200$ square metres. Note that the side parallel to the wall is exactly half the fencing, which is a general result for this type of problem: with one free side, half of the available length goes to the side opposite the wall.

Without the wall, four sides must be fenced: $2x+2y=240$, so $x+y=120$ and $A=x\left(120-x\right)=120x-x^{2}$, maximal at $x=60$, with $y=60$ and $A=3{,}600$ square metres. The wall therefore DOUBLES the available area for the same length of fence, and the optimal shape goes from a square to a rectangle twice as long as it is wide. It is a good example of what a constraint changes: the optimum moves, and not only its value.

First step, IDENTIFY the quantity to optimize and the variables it depends on. Second step, WRITE the constraint that links these variables. Third step, SUBSTITUTE to reduce the objective to a single variable, stating its domain. Fourth step, DIFFERENTIATE, set to zero, check with the second derivative or the sign of the derivative, and also check the ENDPOINTS of the domain, where the optimum can lie when the derivative does not vanish inside.

By the chain rule, $\dfrac{dC}{dt}=C'(q)\times \dfrac{dq}{dt}$. Here $C'(50)=50+20=70$ dollars per unit and $\dfrac{dq}{dt}=3$ units per week, so $\dfrac{dC}{dt}=70\times 3=210$ dollars per week. Take care with the units: dollars per unit times units per week do give dollars per week.

$C'(20)=40$, so $\dfrac{dC}{dt}=40\times 3=120$ dollars per week. The cost therefore grows almost twice as fast at 50 units as at 20 units, although the pace of production is the same. The reason is that the marginal cost itself increases with $q$: this is the dynamic translation of diminishing returns.

$P(q)=R(q)-C(q)=80q-q^{2}-800$, so $P'(q)=80-2q$ and $P'(50)=80-100=-20$ dollars per unit. Then $\dfrac{dP}{dt}=-20\times 3=-60$ dollars per week: profit DECREASES by 60 dollars per week, even though production is rising. The firm has gone past its production optimum.

Profit stops increasing when $P'(q)=0$, that is, at $q=40$. Before that, $\dfrac{dP}{dt}>0$; after, it is negative. Producing more beyond 40 units therefore loses money every week, which is exactly the result of the exercise on maximum profit, seen this time from the point of view of time.

$\overline{C}'(q)=0.5-\dfrac{800}{q^{2}}$, so $\overline{C}'(50)=0.5-\dfrac{800}{2{,}500}=0.5-0.32=0.18$ dollar per unit. Then $\dfrac{d\overline{C}}{dt}=0.18\times 3=0.54$ dollar per week: the average cost rises by about 54 cents per unit each week. The positive sign was predictable, since 50 exceeds the quantity of minimum average cost, which is 40: beyond it, the average cost climbs back.

$R=pq$, so $R'(q)=p+q\dfrac{dp}{dq}=p\left(1+\dfrac{q}{p}\dfrac{dp}{dq}\right)=p\left(1+\dfrac{1}{E}\right)$, since $\dfrac{1}{E}=\dfrac{dp}{dq}\dfrac{q}{p}$. The formula therefore links marginal revenue, price and elasticity directly.

If the demand is INELASTIC, $-1<E<0$, so $\dfrac{1}{E}<-1$ and $1+\dfrac{1}{E}<0$: marginal revenue is NEGATIVE. Selling one more unit therefore lowers total revenue.

At the profit optimum, $R'(q)=C'(q)$. But marginal cost is POSITIVE, since producing one more unit always costs something. Marginal revenue must therefore be positive at the optimum, which requires $1+\dfrac{1}{E}>0$, so $\left|E\right|>1$: the demand is elastic. In other words, in the inelastic range the firm could both increase its revenue and reduce its cost by producing less: it is certainly not at the optimum.

The optimum was at $q=40$, with $p=80$ dollars. Elasticity: $E=\dfrac{-p}{100-p}=\dfrac{-80}{20}=-4$. The absolute value is 4, well above 1: the demand is indeed elastic at the optimum, in line with the general result. We can check the formula of part a): $p\left(1+\dfrac{1}{E}\right)=80\left(1-0.25\right)=60$, which gives back exactly the marginal revenue computed earlier.

That he is describing a missed opportunity rather than a position of strength. If the demand is truly inelastic at the current price, raising the price would make revenue RISE while reducing the quantity produced, and therefore the cost: profit would increase on both sides at once. A firm that finds itself in the inelastic range has simply set its price too low. The position of strength does exist, but it shows up as a low elasticity IN ABSOLUTE VALUE compared with competitors, not as a price chosen in the inelastic range. The practical answer is: raise your price until $\left|E\right|$ exceeds 1, then until $R'=C'$.

Revenue as a function of the price: $R=p\left(900-30p\right)=900p-30p^{2}$. Cost: $C=2\left(900-30p\right)+400=1{,}800-60p+400=2{,}200-60p$. Profit: $P(p)=900p-30p^{2}-2{,}200+60p=960p-30p^{2}-2{,}200$. Then $P'(p)=960-60p$, zero at $p=16$ dollars. Quantity: $q=900-480=420$ meals. Profit: $P(16)=15{,}360-7{,}680-2{,}200=5{,}480$ dollars per day.