The map
A walker's map shows a mountain on a flat sheet of paper. How do you tell from it which way is uphill?
This is the object a walker carries up a mountain, and it is read the same way.
- Crowded contours mean steep ground. Neighbouring contours differ by a fixed rise, so the less horizontal distance between them, the faster the climb.
- Which way is uphill is settled by the labels, not by the shape of the curves.
- Two level curves at different heights never cross. A point cannot be at two heights at once.
Stand at a point and look east: the steepness of the ground in that direction is $f'_{1}.$ Look north: that is $f'_{2}.$ Each partial derivative asks about one step in one direction, and the map answers it directly.
Guess first. The map below is another drawing of the same kind of hill. Read the two signs at $P$ before you read the Method after the question.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
The map shows the height $f(x, y)$.
At the point $P$, what are the signs of $f_1'$ and $f_2'$?
- Why is that answer tempting? Right about east, wrong about north. The summit lies east of P and south of it, so stepping north from P moves onto lower contours.
- Why is that answer tempting? Both signs backwards — the usual result of describing the shape of the picture instead of reading the numbers on it.
- Why is that answer tempting? Reads ‘P is not the summit, so everything falls away from it’. Only the northward step falls. The eastward step climbs, because the high ground is east of P.
A partial derivative asks about one step in one direction, and the map answers it.
$f_1'$: hold $y$, step east. From $P$ that is toward the high ground, so $f_1' > 0$.
$f_2'$: hold $x$, step north. That is away from the summit, onto lower contours, so $f_2' < 0$.
At $Q$ both answers come out the other way round — same reading, other side of the hill.
Two signs, read off the map
On the map above, what are the signs of $f'_{1}$ and $f'_{2}$ at $P$?
The map is the one in the question above. The notes draw another hill, with contours labelled 7, 8 and 6 where this one has 8, 9, 6.5 and 4.5; the reading is the same.
Try it yourself first. Then open one step at a time.
Slide east from $P{:}$ the contours crossed are labelled $8,$ then $9$ — the ground climbs, so $f'_{1}>0.$ Slide north from $P{:}$ the contours fall away to $6.5$ and $4.5,$ so $f'_{2}<0.$
Note what was not used. Nothing about the shape of the ellipses entered the argument; only the labels did. The high ground here lies to the east and to the south of $P,$ and each partial derivative picked up exactly the one direction it asks about.
$f'_{1}>0$ and $f'_{2}<0$ at $P.$
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
On a contour map, put a finger on a point $R.$ Sliding it a little to the right, you cross the labels $5$ and then $4.$ Sliding it up from $R,$ you cross $6$ and then $7.$ What are the signs of the partial derivatives at $R$?
- Why is that answer tempting? One slide climbs, and the other is taken to climb as well. Each slide answers its own question: going right the labels fall.
- Why is that answer tempting? The two slides are swapped: the slide to the right answers $f'_{1},$ the slide up answers $f'_{2}.$
- Why is that answer tempting? The labels are all Method 9.3 needs. Neither answer depends on the shape of the curves (Watch out 9.5).
Going right the labels fall, from $5$ to $4{:}$ $f'_{1}<0.$ Going up they rise, from $6$ to $7{:}$ $f'_{2}>0.$ Two slides, two numbers each, and nothing about the shape of the curves.
What economists actually draw
Every flat two-variable diagram you have seen in economics is a contour map. Of what?
An isoquant is a level curve of a production function: the input combinations giving the same output, $F(K,L)=Y_{0}.$ An indifference curve is a level curve of a utility function: the bundles a consumer likes equally, $U(x,y)=u_{0}.$ Higher curve means more output, or more satisfaction, and that convention is the only thing added.
Every flat two-variable diagram in microeconomics is one of these. The surface is never drawn because it never needs to be.
The shape carries economics too. A Cobb–Douglas isoquant bends towards the origin, which says that capital can be replaced by labour, but at a rate that gets worse the further the substitution is pushed. Walking along a contour computes that rate exactly.
Recap Production, utility and the margin
| word | what it is |
|---|---|
| production function $F(K,L)$ | output from capital $K$ and labour $L$ |
| utility function $U(x,y)$ | how much a consumer likes the bundle of $x$ units of one good and $y$ of another |
| marginal product of capital | $F'_{K},$ the extra output from one more unit of capital, labour held fixed |
| complements | inputs such that more of one raises the marginal product of the other |
| budget line | the bundles that cost exactly the income (Lecture 1) |
Guess first. The example after the question computes the three signs that matter for a production function. Say what the second one means before you read it.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
For a production function $F(K, L)$ it is found that $$F_K' > 0 \qquad\text{and}\qquad F_{KK}'' < 0$$
What does the second of these say?
- Why is that answer tempting? Reads the second derivative as though it were the first. $F_K'>0$ has already settled that output rises; $F_{KK}''$ is about the rate at which it does.
- Why is that answer tempting? That is a claim about the cross partial $F_{KL}'',$ a different object. $F_{KL}''>0$ is what says more capital raises the marginal product of labour.
- Why is that answer tempting? Compares two different inputs. $F_{KK}''$ involves capital only and says nothing whatever about labour.
$F_K'$ is the marginal product of capital — the extra output from one more unit. $F_{KK}''$ is the derivative of that, so it says how the marginal product changes as capital grows.
Negative means the marginal product is falling. Output still rises; it just rises by less each time. That is diminishing marginal product, and for $F = AK^{a}L^{b}$ it holds exactly when $a < 1$.
The signs are the economics
For $F(K,L)=AK^{a}L^{b}$ with $A,a,b>0,$ compute $F'_{K},$ $F''_{KK}$ and $F''_{KL},$ and say what each sign asserts.
Try it yourself first. Then open one step at a time.
Differentiating, $$ \begin{aligned} F'_{K} &= aAK^{a-1}L^{b}\,,\\ F''_{KK} &= a(a-1)AK^{a-2}L^{b}\,,\\ F''_{KL} &= abAK^{a-1}L^{b-1}\,. \end{aligned} $$
The first is positive: more capital, more output — $F'_{K}$ is the marginal product of capital. The second is negative exactly when $a<1{:}$ the marginal product is falling, so output rises at a diminishing rate.
The third is positive: more capital raises the marginal product of labour, which is what it means to call capital and labour complements. By Fact 8.21 that last statement is automatically symmetric — more labour raises the marginal product of capital by the same amount — which is not obvious in advance and is a theorem doing real work.
$F'_{K}>0$ (more capital, more output); $F''_{KK}<0$ exactly when $a<1$ (diminishing marginal product); $F''_{KL}>0$ (capital and labour are complements).
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
A firm's output is $F(K,L)=4K^{1.4}L^{0.5}.$ Then $$ F''_{KK}=2.24\,K^{-0.6}L^{0.5}>0\,. $$ What does that sign say?
- Why is that answer tempting? That would be $F'_{K}<0.$ Here $F'_{K}=5.6K^{0.4}L^{0.5}>0{:}$ output rises with capital.
- Why is that answer tempting? Diminishing marginal product needs $a<1.$ With $a=1.4$ the same formula asserts the opposite: Watch out 9.8.
- Why is that answer tempting? Complements is a statement about the cross partial $F''_{KL},$ a different object. Here $F''_{KL}=2.8K^{0.4}L^{-0.5}>0,$ so they are complements.
$F'_{K}=5.6K^{0.4}L^{0.5}$ is the marginal product of capital, and $F''_{KK}$ is its derivative. Positive means the marginal product is rising: each unit of capital adds more than the last. That is the economy of Watch out 9.8, with $a=1.4>1.$
More variables
In one variable, “$x$ is close to $a$” meant $|x-a|$ small. What does close mean for a bundle of ten inputs?
Three inputs, or ten. Nothing conceptual changes: there is one partial derivative per variable, each found by holding the other $n-1$ still. What is lost is the picture, and only the picture.
What does need replacing is the phrase “$x$ is close to $a$”, which in one variable meant $|x-a|$ small.
With that, continuity means in $n$ variables exactly what it meant in Lecture 5: a small enough change in the inputs makes only a small change in the output. Only the meaning of “small” had to be supplied: a small distance.
The general Cobb–Douglas function $F = Ax_{1}^{a_{1}}\cdots x_{n}^{a_{n}}$ is worth one remark. Taking logarithms, $$ \ln F = \ln A + a_{1}\ln x_{1} + \cdots + a_{n}\ln x_{n}\,, $$ which is linear. That is Lecture 2's trick again — take logarithms and a multiplicative thing becomes additive — and it is a large part of why this functional form is estimated so often: in logarithms it is a straight line.
Close, in four variables, and straight in logarithms
This example is the app's; nothing in it is new. A firm uses capital, labour, energy and materials.
(a) Last year it used the bundle $(10,4,7,2)$ of the four inputs, this year $(12,8,6,4).$ How far apart are the two bundles?
(b) Its output is $F=3x_{1}^{0.5}x_{2}^{0.3}x_{3}^{0.1}x_{4}^{0.1}.$ Write $\ln F$ as a linear expression, and use it to say by about how much output rises when $x_{1}$ alone rises by $1\%.$
Try it yourself first. Then open one step at a time.
(a) The differences, squared. The coordinates differ by $2,$ $4,$ $-1$ and $2,$ so by Definition 9.9 $$ \begin{aligned} &\sqrt{2^{2}+4^{2}+(-1)^{2}+2^{2}}\\ &=\sqrt{4+16+1+4}\\ &=\sqrt{25}=5\,. \end{aligned} $$ The sign of a difference does not matter: it is squared.
(b) Logarithms. By the rules of Lecture 2, $$ \begin{aligned} \ln F&=\ln3+0.5\ln x_{1}\\ &\quad+0.3\ln x_{2}+0.1\ln x_{3}\\ &\quad+0.1\ln x_{4}\,. \end{aligned} $$
A rise of $1\%$ in $x_{1}.$ Then $\ln x_{1}$ grows by $\ln1.01\approx0.01$ (the linear approximation $\ln(1+h)\approx h$), and $\ln F$ by $0.5\cdot0.01=0.005.$ So $F$ is multiplied by about $e^{0.005}\approx1.005{:}$ output rises by about $0.5\%.$ The exponent is the percentage effect, which is the elasticity met in Lecture 8.
(a) $5.$
(b) $\ln F=\ln3+0.5\ln x_{1}+0.3\ln x_{2}+0.1\ln x_{3}+0.1\ln x_{4};$ output rises by about $0.5\%.$
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
What is the distance between the points $(1,2,3)$ and $(3,5,9)$?
- Why is that answer tempting? Adds the differences $2+3+6$ instead of using Pythagoras. That is the length of a walk along the edges of the box, not across it.
- Why is that answer tempting? The sum of the squares, $4+9+36,$ with the square root left out.
- Why is that answer tempting? Puts the differences under the root without squaring them.
The differences are $2,$ $3$ and $6,$ so by Definition 9.9 the distance is $\sqrt{4+9+36}=\sqrt{49}=7.$
Walking across the map
You walk across a hill, east and north at once. How fast are you climbing?
Now let both variables move together, say with time: $x=g(t)$ and $y=h(t).$ Then $z = f\bigl(g(t),h(t)\bigr)$ depends on $t$ alone, and we can ask how fast it changes.
If $f$ is differentiable and $x=g(t)$, $y=h(t)$ are differentiable, then $$ \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}\,. \tag{9.1} $$ One contribution per variable, and they add.
Read as a walk across the landscape, (9.1) says $$ \begin{aligned} &\text{rate of climb}\\ &= (\text{steepness east})\times(\text{speed east})\\ &\quad+ (\text{steepness north})\\ &\qquad\times(\text{speed north})\,, \end{aligned} $$ which is what anybody would guess. The units make it unforgettable and rule out every other pairing: output per extra machine times machines per year, plus output per extra worker times workers per year, is output per year. If the partials are ever paired with the wrong rates, the units catch it at once. It is the chain rule of Lecture 3 with a second term.
Guess first. The example after the question has the same numbers.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
A firm's output is $Y = F(K, L).$ Right now $$ \begin{aligned} &F_K' = 4\,, \quad F_L' = 6\,,\\ &\frac{dK}{dt} = 3\,, \quad \frac{dL}{dt} = -2\,. \end{aligned} $$
Capital is being built up; the workforce is shrinking. How fast is output changing?
- Why is that answer tempting? $4\cdot3+6\cdot2,$ with the minus sign dropped. A shrinking workforce subtracts from output — the sign is the entire content of the second term.
- Why is that answer tempting? Counts capital and forgets labour. The chain rule has one term per variable and both variables are moving.
- Why is that answer tempting? $4\cdot(-2)+6\cdot3$ — each partial paired with the wrong rate. $F_K'$ goes with $dK/dt,$ always; the units make it impossible to pair them any other way.
Output is momentarily unchanged.
Notice what that does not mean. Nothing here is idle: capital accumulation is adding twelve units a year and the shrinking workforce is taking twelve away. They happen to cancel at this instant, and a moment later they will not.
Capital up, labour down
A firm's output is $Y=F(K,L).$ At this moment $F'_{K}=4,$ $F'_{L}=6,$ $dK/dt=3$ and $dL/dt=-2{:}$ capital is being built up and the workforce is shrinking. How fast is output changing?
Try it yourself first. Then open one step at a time.
By (9.1), $$ \begin{aligned} \frac{dY}{dt} &= F'_{K}\frac{dK}{dt} + F'_{L}\frac{dL}{dt}\\ &= 4(3) + 6(-2)\\ &= 12-12 = 0\,. \end{aligned} $$
Output is momentarily unchanged — and notice what that does not mean. Nothing here is idle: capital accumulation is adding twelve units a year and the shrinking workforce is removing twelve. They happen to cancel at this instant, and a moment later they will not.
There is a free check available whenever $g$ and $h$ are known explicitly: substitute them into $f$ first and differentiate the resulting function of $t$ alone. The two routes must agree, and running both is worth the time while the rule is new.
$dY/dt=0{:}$ at this instant output is unchanged, because the $+12$ from capital and the $-12$ from labour cancel.
Two routes, one answer
This example is the app's: the free check of Example 9.12, carried out. Let $z=xy^{2},$ and walk along $x=t^{2},$ $y=3-t.$ Find $dz/dt$ at $t=1$ in both ways.
Try it yourself first. Then open one step at a time.
By the chain rule. At $t=1$ we are at $x=1,$ $y=2.$ The partials are $\dfrac{\partial z}{\partial x}=y^{2}=4$ and $\dfrac{\partial z}{\partial y}=2xy=4;$ the speeds are $\dfrac{dx}{dt}=2t=2$ and $\dfrac{dy}{dt}=-1.$ By (9.1), $$ \frac{dz}{dt}=4\cdot2+4\cdot(-1)=4\,. $$
Substituting first. Along the path $z=t^{2}(3-t)^{2},$ a function of $t$ alone. By the product rule and the chain rule of Lecture 3, $$ \begin{aligned} \frac{dz}{dt}&=2t(3-t)^{2}-2t^{2}(3-t)\,, \end{aligned} $$ which at $t=1$ is $2\cdot4-2\cdot2=4.$ The two routes agree.
$dz/dt=4$ at $t=1,$ by both routes.
In applied work the rule is more often wanted in relative terms, because economists quote growth rates rather than levels.
Dividing (9.1) through by $z$, $$ \frac{\dot z}{z} = \mathrm{El}_{x}z\cdot\frac{\dot x}{x} + \mathrm{El}_{y}z\cdot\frac{\dot y}{y}\,, $$ where a dot denotes $d/dt$. The growth rate of $z$ is the elasticity-weighted sum of the growth rates of $x$ and $y$.
Divide each side of (9.1) by $z$ and multiply and divide the first term by $x$ and the second by $y{:}$ $$ \begin{aligned} \frac{1}{z}\frac{dz}{dt} &= \frac{x}{z}\frac{\partial z}{\partial x}\cdot\frac{1}{x}\frac{dx}{dt}\\ &\quad+ \frac{y}{z}\frac{\partial z}{\partial y}\cdot\frac{1}{y}\frac{dy}{dt}\,, \end{aligned} $$ which is the statement. It is also Lecture 3's relative-rate identity with a second variable added.
So if demand $D(p,m)$ has price elasticity $-0.8$ and income elasticity $1.2,$ and prices rise $3\%$ a year while incomes rise $2\%,$ then demand grows at about $-0.8(3\%) + 1.2(2\%) = 0.$ For Cobb–Douglas the elasticities are the exponents, as Lecture 8 showed, so the arithmetic is immediate. A partial elasticity turns a rate in units into a rate in per cent, which is the relative rate of change of Lecture 3 again.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
Output is $Y=K^{0.3}L^{0.7}.$ Capital grows by $5\%$ a year and labour by $1\%$ a year. By about how much does output grow per year?
- Why is that answer tempting? Adds the two growth rates. Fact 9.14 weights each one by its elasticity, here $0.3$ and $0.7.$
- Why is that answer tempting? The weights are swapped: $0.7\cdot5+0.3\cdot1.$ The elasticity with respect to $K$ is the exponent of $K,$ which is $0.3.$
- Why is that answer tempting? The plain average of $5\%$ and $1\%.$ The weights are the elasticities, not one half each.
For Cobb–Douglas the elasticities are the exponents: $\mathrm{El}_{K}Y=0.3$ and $\mathrm{El}_{L}Y=0.7.$ By Fact 9.14, $$ \begin{aligned} \frac{\dot Y}{Y}&=0.3\cdot5\%+0.7\cdot1\%\\ &=1.5\%+0.7\%=2.2\%\,. \end{aligned} $$
Walking along a contour
Is there a direction in which you can walk on a hillside without climbing or descending at all?
One direction of travel is special. Walk along a level curve and, by definition, the height does not change at all.
Choose a direction and walk
Left, the hill as a map in shaded bands (every edge of a band is a level curve) with a straight walk from the cross; right, your height along the walk.
- Turn the direction dial slowly. Under the graphs, compare the sum of the two chain-rule terms $f_{1}\cdot dx/dt$ and $f_{2}\cdot dy/dt$ with the slope of the graph. They agree every time: that is (9.1), measured.
- Before you press anything else, find by turning the dial a direction in which the right-hand graph comes out flat. There are two, opposite to each other.
- Press Find the flat direction. You are walking along a level curve, and the two terms cancel. That cancellation is the Fact below.
- Press the button with the function's name until it reads $F(K,L)=10\sqrt{KL},$ and find the flat direction again: it runs along an isoquant.
Along a level curve of $f$, wherever $f'_{2}\ne0$, $$ \frac{dy}{dx} = -\,\frac{f'_{1}}{f'_{2}}\,. \tag{9.2} $$
Walking along the curve means $z$ is constant, so $dz/dt=0,$ and (9.1) becomes $$ f'_{1}\frac{dx}{dt} + f'_{2}\frac{dy}{dt} = 0\,. $$ Dividing by $f'_{2}\,dx/dt$ gives (9.2).
That single line is the most used formula in microeconomics. On an indifference curve it is the marginal rate of substitution: the rate at which the consumer is willing to give up $y$ for $x$ while staying equally happy. On an isoquant it is the marginal rate of technical substitution.
It is also implicit differentiation from Lecture 4 in one line: the equation $f(x,y)=c$ defines $y$ implicitly near a point where $f'_{2}\ne0,$ and (9.2) is its derivative.
Guess first. Use (9.2) on the numbers in the question, and check the answer against your sense of the curve.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
At a point, $f_1' = 3$ and $f_2' = 6$.
What is the slope $\dfrac{dy}{dx}$ of the level curve through that point?
- Why is that answer tempting? The ratio upside down: −f₂′/f₁′ instead of −f₁′/f₂′. The sense check kills it — y is twice as productive here, so you need less of it, and the curve should be shallow.
- Why is that answer tempting? The right ratio with no minus sign. A level curve of an increasing function slopes down: more of one variable has to mean less of the other.
- Why is that answer tempting? Inverted and unsigned at once — the formula recalled as a bare ratio of the two partials, with neither the order nor the sign attached.
Along the curve $z$ does not change, so $$ \begin{aligned} &3\,dx + 6\,dy = 0\\ &\Longrightarrow\quad \frac{dy}{dx} = -\frac{3}{6} = -\frac12 \end{aligned} $$
In plain numbers: two extra units of $x$ raise $z$ by $6,$ and one fewer unit of $y$ lowers it by $6.$ They cancel, so the step $(+2, -1)$ keeps you on the curve — a slope of $-\tfrac12.$
Three against six
At a point, $f'_{1}=3$ and $f'_{2}=6.$ What is the slope of the level curve there?
Try it yourself first. Then open one step at a time.
By (9.2), $dy/dx = -3/6 = -\tfrac12.$
In plain numbers: two extra units of $x$ raise $z$ by $6,$ and one fewer unit of $y$ lowers it by $6.$ They cancel exactly, so the step $(+2,-1)$ keeps you on the curve — a slope of $-\tfrac12,$ and $y,$ being twice as productive here, is indeed the one you need less of.
The Cobb–Douglas trade-off
Find the slope of the indifference curves of $U(x,y)=x^{a}y^{b}.$
Try it yourself first. Then open one step at a time.
As in Lecture 8, $U'_{1}=aU/x$ and $U'_{2}=bU/y,$ so by (9.2) $$ \begin{aligned} \frac{dy}{dx} &= -\frac{aU/x}{bU/y}\\ &= -\frac{a}{b}\cdot\frac{y}{x}\,. \end{aligned} $$ Two things fall out of that, and the second is the deeper one.
First, the trade-off depends only on the ratio $y/x$ and the two exponents, so it is constant along any ray from the origin: doubling both goods leaves it unchanged.
Second, $U$ cancelled. So any increasing transform of $U$ — $U^{2},$ $\ln U,$ $5U+3$ — has exactly the same level curves and exactly the same trade-off at every point. The numbers a utility function produces therefore carry no information at all; only the level curves do. That is twenty seconds of cancellation and one of the foundational ideas in consumer theory.
$\dfrac{dy}{dx}=-\dfrac{a}{b}\cdot\dfrac{y}{x}{:}$ it depends only on the ratio $y/x,$ and $U$ itself has cancelled.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
A consumer has utility $U(x,y)=x^{2}y^{3}.$ What is the slope $dy/dx$ of the indifference curve through the point $(3,4)$?
- Why is that answer tempting? The ratio upside down: $-U'_{2}/U'_{1}.$ Here $U'_{1}=2xy^{3}=384$ and $U'_{2}=3x^{2}y^{2}=432;$ $y$ has the bigger partial, so you need less of it and the curve is shallow.
- Why is that answer tempting? The quantities upside down: $-\frac{a}{b}\cdot\frac{x}{y}=-\frac23\cdot\frac34.$ In Example 9.18 the ratio is $y/x.$
- Why is that answer tempting? The right size without the minus sign. Both partials are positive, so more $x$ has to be paid for with less $y{:}$ the curve slopes down.
By Example 9.18 with $a=2,$ $b=3,$ $$ \begin{aligned} \frac{dy}{dx}&=-\frac{2}{3}\cdot\frac{4}{3}=-\frac{8}{9}\,. \end{aligned} $$ Directly: $-U'_{1}/U'_{2}=-384/432=-\tfrac89.$
The best point on a line
More is always better, so utility has no maximum. But on a budget line, which bundle is best?
The map also answers a question we have no algebra for yet.
Guess first. Slide a finger along the budget line in the picture, and watch which indifference curves you cross.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
You may choose any point on the budget line. Higher curves are better.
Which of $P$, $Q$, $R$, $S$ is best?
- Why is that answer tempting? The end of the line is not special. P sits on a low curve, and sliding to the right crosses onto higher ones.
- Why is that answer tempting? Close, and wrong for the reason that is the whole point: at Q the line crosses a curve, so sliding along it one way must reach a higher one.
- Why is that answer tempting? Past the best point and on the way back down. Value rises along the line and then falls; S is on the far side of the turn.
At $P$, $Q$ and $S$ the line crosses a curve — and wherever it crosses, sliding along it one way reaches a higher curve.
At $R$ it touches. There is nowhere left to slide.
Touching means equal slopes, so $-U_1'/U_2' = -p/q$: the central condition of consumer theory, read off a picture.
The best point on a line
A consumer has $12$ to spend; good $x$ costs $2$ and good $y$ costs $3,$ so the affordable bundles are those on $2x+3y=12.$ Utility is $U(x,y)=xy.$ Which affordable bundle is best?
Try it yourself first. Then open one step at a time.
This is not “maximise $U$”, which has no answer since more is always better; it is the best point on a line, which is a different question with a real one. Substituting $y=(12-2x)/3,$ $$ \begin{aligned} U &= \frac{x(12-2x)}{3}\,,\\ \frac{dU}{dx} &= \frac{12-4x}{3}\,, \end{aligned} $$ which is $0$ at $x=3;$ so $x=3,$ $y=2$ and $U=6{:}$ half the income spent on each good.
Now read it off the map instead. At any point where the budget line crosses an indifference curve, sliding along the line one way or the other reaches a higher curve, so no such point can be best. At the best point the line must touch a curve — and touching means equal slopes: $$ \begin{aligned} &-\frac{U'_{1}}{U'_{2}} = -\frac{p}{q}\,,\\ &\text{here}\quad -\frac{y}{x} = -\frac{2}{3}\,, \end{aligned} $$ which with $2x+3y=12$ gives $x=3,$ $y=2$ again. In words: at the best bundle the rate at which the consumer is willing to trade equals the rate at which the market lets them.
$x=3,$ $y=2,$ with $U=6{:}$ the point where the budget line touches the indifference curve $xy=6.$
That tangency condition is the central result of consumer theory, and it has just been obtained by sliding a finger along a line. Lecture 11 derives it properly, for constraints of any shape, by the Lagrange multiplier method; the hard part of the idea is already done.
What you should take away
- A level curve is where $f(x,y)=c;$ several make a contour map. Crowded contours mean steep ground, the labels say which way is up, and curves at different heights never cross. To sign a partial derivative, take one step in one direction and read the labels.
- Isoquants and indifference curves are level curves, which is why every flat two-variable diagram in microeconomics works.
- A model is a functional form plus sign assumptions. $F'_{K}>0,$ $F''_{KK}<0$ and $F''_{KL}>0$ say: more capital raises output, at a diminishing rate, and capital and labour are complements.
- In $n$ variables nothing changes but the picture: one partial per variable, and $\|\mathbf{x}-\mathbf{a}\|$ replaces $|x-a|.$
- Chain rule: $\dfrac{dz}{dt} = \dfrac{\partial z}{\partial x} \dfrac{dx}{dt} + \dfrac{\partial z}{\partial y}\dfrac{dy}{dt}$ — steepness times speed, one term per variable, added. In relative terms the weights are the elasticities.
- Walking along a level curve gives $dz/dt=0,$ hence $\dfrac{dy}{dx} = -\dfrac{f'_{1}}{f'_{2}}{:}$ the marginal rate of substitution. The variable with the bigger partial is the one you need less of.
- For $U=x^{a}y^{b}$ the trade-off is $-(a/b)(y/x),$ which depends only on the ratio of the goods — and $U$ cancels, so only the level curves of a utility function carry information, never its values.
- The best affordable bundle is where the budget line touches an indifference curve: $-U'_{1}/U'_{2} = -p/q.$ Where it crosses, you can always do better.
Think further
Questions without computation. Think on your own or with a classmate before opening the thought.
Discussion question 1 — the other side of the summit
On the contour map above, what are the signs of $f'_{1}$ and $f'_{2}$ at a point placed symmetrically to $P$ on the far side of the summit?
One way to think about it
Both signs turn round: $f'_{1}<0$ and $f'_{2}>0.$ Every contour of this hill is an ellipse centred on the summit, so the map looks the same after a half-turn about the summit. A step east from the mirror point corresponds to a step west from $P,$ which goes down; a step north corresponds to a step south from $P,$ which goes up. The point $Q$ in the map of the first question lies on that side of the hill, and the answer figure after Example 9.4 shows its two signs.
Discussion question 2 — rising, but more slowly
For a production function it is found that $F'_{K}>0$ and $F''_{KK}<0.$ What does the second of these say?
(A) output falls as capital rises;
(B) capital and labour are complements;
(C) capital is less productive than labour;
(D) output rises, but each extra unit adds less than the last.
One way to think about it
(D). $F''_{KK}$ is the derivative of the marginal product $F'_{K},$ so negative means the marginal product is falling: output still rises, by less each time (Watch out 9.7). (A) reads the second derivative as the first; (B) is a statement about the cross partial $F''_{KL};$ (C) compares two inputs, and $F''_{KK}$ involves capital only.
Discussion question 3 — cancelling is not idle
Right now $F'_{K}=4,$ $F'_{L}=6,$ $dK/dt=3$ and $dL/dt=-2.$ How fast is output changing, and why is “nothing is happening” the wrong description of the answer?
One way to think about it
$dY/dt=4\cdot3+6\cdot(-2)=12-12=0$ (Example 9.12). But two things are happening at once: capital accumulation adds $12$ units a year and the shrinking workforce removes $12.$ They cancel at this instant only; as $K$ and $L$ change, so do the partials, and a moment later the two terms will not cancel. A derivative of zero is a statement about one instant, not a steady state.
Discussion question 4 — the sense check at work
At a point, $f'_{1}=3$ and $f'_{2}=6.$ The slope of the level curve is
(A) $-\tfrac12;$
(B) $-2;$
(C) $\tfrac12;$
(D) $2.$
Which of the wrong answers does the sense check of Warning 9.16 eliminate immediately?
One way to think about it
The answer is (A), $-3/6=-\tfrac12.$ The sense check — the variable with the bigger partial is the one you need less of — says that $y,$ with partial $6,$ is needed less, so the curve is shallow: the slope lies between $-1$ and $1.$ That eliminates (B) $-2$ and (D) $2,$ the inverted ratios. (C) $\tfrac12$ passes the size test; the other half of Watch out 9.16, the minus sign, removes it: with both partials positive, more $x$ must be paid for with less $y.$
Discussion question 5 — the same consumer
Why do $U=xy$ and $U=\ln x+\ln y$ describe exactly the same consumer?
One way to think about it
$\ln x+\ln y=\ln(xy),$ an increasing function of $xy.$ So the two utility functions rank every pair of bundles the same way and have the same level curves: $xy=c$ is the curve $\ln x+\ln y=\ln c.$ The trade-offs agree too. For $U=xy,$ $-U'_{1}/U'_{2}=-y/x;$ for $\ln x+\ln y,$ $-\frac{1/x}{1/y}=-y/x.$ Only the labels of the curves differ, and by Example 9.18 the labels carry no information.
Discussion question 6 — the price of x doubles
In Example 9.19, what happens to the best bundle if the price of $x$ doubles? Answer from the tangency condition rather than by redoing the substitution.
One way to think about it
The price of $x$ becomes $4,$ so the budget line is $4x+3y=12$ and the tangency condition is $-y/x=-4/3,$ that is $y=\tfrac43x.$ Then $4x+4x=12,$ so $x=\tfrac32$ and $y=2.$ The amount of $x$ halves and $y$ is unchanged: the consumer still spends half the income, $6,$ on each good. With $U=xy$ that is always so.
Discussion question 7 — two contours that cross
Two level curves of the same function, at heights $6$ and $7,$ are drawn crossing each other. What has gone wrong?
One way to think about it
The drawing. At the crossing point the function would have the value $6$ (it lies on the first curve) and the value $7$ (it lies on the second), and a function has one value at each point (Fact 9.2, item 3). So either a curve is mislabelled, or the two curves belong to different functions.