Contour maps, and walking on them

(a) The two level curves. On $f=4$ with $y=0$ we get $x^{2}=4,$ so $x=\pm2;$ with $x=0,$ $4y^{2}=4,$ so $y=\pm1.$ On $f=16{:}$ $x=\pm4$ and $y=\pm2.$ Both are ellipses centred at the origin, twice as wide as they are tall. (The origin is the lowest point, $f(0,0)=0.$)

(b) Where the map is crowded. Along the $x$-axis the two contours lie at $x=2$ and $x=4,$ a distance $2$ apart; along the $y$-axis at $y=1$ and $y=2,$ a distance $1$ apart. The same rise, $12,$ over half the distance: by Fact 9.2 the ground is steeper in the $y$-direction.

(c) The slides at $(2,0).$ Sliding right from $(2,0)$ moves outwards, onto higher labels: $f'_{1}>0.$ Sliding up, the value $4+4y^{2}$ rises — but it rises sliding down as well. A slide that climbs whichever way it goes starts at the lowest point of that line: along $x=2$ the lowest point is $(2,0),$ and there $f'_{2}=0.$ The partials agree: $f'_{1}=2x=4>0$ and $f'_{2}=8y=0.$

At $(0,1).$ Sliding up moves outwards: $f'_{2}>0.$ Sliding right the labels rise, and sliding left they rise too: $(0,1)$ is the lowest point of the line $y=1,$ so $f'_{1}=0.$ Indeed $f'_{1}=2x=0$ and $f'_{2}=8y=8>0.$

(a) The marginal product. With $L$ fixed, $F'_{K}=12+L-2K.$ For $L=4$ this is $16-2K,$ positive exactly when $K<8.$ Beyond $K=8$ more capital lowers output (with four workers): this formula does not assert $F'_{K}>0$ everywhere.

(b) The second derivatives. $F''_{KK}=-2<0{:}$ the marginal product of capital falls as capital grows — output rises more slowly, it does not fall (Watch out 9.7). $F''_{KL}=1>0{:}$ more labour raises the marginal product of capital, so capital and labour are complements.

(c) The other order. $F'_{L}=K,$ so $F''_{LK}=1,$ the same number. More capital raises the marginal product of labour by exactly as much. The order of differentiation does not matter, which is Fact 8.21.

(a) The distance. The differences are $4,$ $-2$ and $4,$ so $$ \begin{aligned} &\sqrt{4^{2}+(-2)^{2}+4^{2}}\\ &=\sqrt{16+4+16}=\sqrt{36}=6\,. \end{aligned} $$

Closer to the origin. $\sqrt{1+16+4}=\sqrt{21}\approx4.58$ for the first point, and $\sqrt{9+4+36}=\sqrt{49}=7$ for the second. The first is closer.

(b) Logarithms. $\ln F=\ln5+0.6\ln x_{1}+0.4\ln x_{2}.$ Doubling both inputs adds $\ln2$ to each logarithm, so $\ln F$ grows by $0.6\ln2+0.4\ln2=\ln2.$ Hence $F$ is multiplied by $e^{\ln2}=2{:}$ output doubles, because the exponents add up to $1.$

(a) The elasticities. $Y'_{K}=0.4AK^{-0.6}L^{0.6}=0.4Y/K,$ so $$\mathrm{El}_{K}Y=\frac{K}{Y}\cdot\frac{0.4Y}{K}=0.4\,,$$ and in the same way $\mathrm{El}_{L}Y=0.6.$ The exponents.

(b) The growth of $Y.$ By Fact 9.14, $$ \begin{aligned} \frac{\dot Y}{Y}&=0.4\cdot3\%+0.6\cdot0.5\%\\ &=1.2\%+0.3\%=1.5\%\,. \end{aligned} $$

(c) Per worker. By Fact 3.15 the relative rate of a quotient is the difference of the relative rates: $$ \begin{aligned} \frac{\dot Y}{Y}-\frac{\dot L}{L}&=1.5\%-0.5\%\\ &=1\%\,. \end{aligned} $$ Equivalently, $Y/L=A(K/L)^{0.4},$ and $K/L$ grows by $3\%-0.5\%=2.5\%,$ of which $0.4\cdot2.5\%=1\%.$

(a) The chain rule. With $\dfrac{dx}{dt}=1$ and $\dfrac{dy}{dt}=-2t,$ $$ \begin{aligned} \frac{dz}{dt}&=y\cdot1+x\cdot(-2t)\\ &=(6-t^{2})-2t^{2}=6-3t^{2}\,. \end{aligned} $$

(b) Substituting first. Along the path $z=t(6-t^{2})=6t-t^{3},$ and $dz/dt=6-3t^{2}.$ The same.

(c) Where it levels off. $6-3t^{2}=0$ gives $t=\sqrt2$ ($t>0$). There $x=\sqrt2,$ $y=6-2=4$ and $z=4\sqrt2\approx5.66.$ Since $dz/dt=6-3t^{2}$ is positive for $t<\sqrt2$ and negative for $t>\sqrt2,$ the height rises and then falls: $t=\sqrt2$ is the highest point of the walk (Fact 2.23 on the function $6t-t^{3}$).

(d) Along the level curve. At $t=\sqrt2$ the path has slope $$\frac{dy/dt}{dx/dt}=\frac{-2\sqrt2}{1}=-2\sqrt2\,.$$ The level curve $xy=4\sqrt2$ through $(\sqrt2,4)$ has slope $$-\frac{f'_{1}}{f'_{2}}=-\frac{y}{x}=-\frac{4}{\sqrt2}=-2\sqrt2\,.$$ They agree: at the instant the climb stops, the walk runs along the contour. That is $dz/dt=0$ in (9.1) read as a picture, the argument of Fact 9.15 backwards.

(a) The slopes. For $U{:}$ $-\dfrac{y}{x}.$ For $V{:}$ $-\dfrac{1/x}{1/y}=-\dfrac{y}{x}.$ For $W{:}$ $W'_{1}=2xy^{2},$ $W'_{2}=2x^{2}y,$ so $-\dfrac{2xy^{2}}{2x^{2}y}=-\dfrac{y}{x}.$ For $Z{:}$ $-\dfrac{-y}{-x}=-\dfrac{y}{x}.$ The same, as in Example 9.18, where $U$ cancelled.

(b) The same order. $V=\ln(xy)$ and $W=(xy)^{2}+1$ are increasing functions of $xy$ (for $xy>0$). So if one bundle has a larger $xy$ than another, it also has a larger $V$ and a larger $W{:}$ the three rank every pair of bundles alike, and the level curve $xy=c$ is the level curve $V=\ln c$ and $W=c^{2}+1.$

(c) Not the same consumer. $Z=-xy$ is a decreasing function of $xy.$ Its level curves are the same hyperbolas $xy=c,$ but the labels run the other way: $Z(1,1)=-1$ and $Z(2,2)=-4,$ so this consumer prefers $(1,1)$ to $(2,2),$ less of both goods. So the curves alone are not the whole information: their order — which way is up, read from the labels as in Fact 9.2 — counts too. What carries no information is the size of the numbers, as long as the order is kept.

(a) Tangency. The slope of the indifference curve is $-\dfrac{2xy}{x^{2}}=-\dfrac{2y}{x},$ the slope of the budget line $-\dfrac{2}{1}.$ Equal slopes: $\dfrac{2y}{x}=2,$ so $y=x.$ With $2x+y=30{:}$ $3x=30,$ $x=10,$ $y=10,$ and $U=100\cdot10=1000.$

(b) Substituting. On the line $y=30-2x$ with $0\le x\le15,$ $$ \begin{aligned} u(x)&=x^{2}(30-2x)\\ &=30x^{2}-2x^{3}\,,\\ u'(x)&=60x-6x^{2}\\ &=6x(10-x)\,. \end{aligned} $$ $u'>0$ on $(0,10)$ and $u'<0$ on $(10,15),$ so $x=10$ gives the largest value, $u(10)=1000;$ at the ends $u(0)=u(15)=0.$

(c) The shares. $2\cdot10=20$ of the $30$ go to $x$ and $10$ to $y{:}$ two thirds and one third — the exponents $2$ and $1$ divided by their sum.