IfF (K, L)denotes the number of units produced whenK units of capital andLunits of labour are used as inputs, economists often ask: What happens to production if we double the inputs of both capital and labour? Will production rise by more or less than a factor of 2? To answer such and related questions, we introduce the following new concept of homogeneityfor functions of two variables.
A functionf of two variablesxandydefined in a domainDis said to behomogeneous of degreekif, for all(x, y)inD,
f (t x, ty)=tkf (x, y) for allt >0 (1) Multiplying both variables by a positive factortwill thus multiply the value of the function by the factortk.
The degree of homogeneity of a function can be an arbitrary number—positive, zero, or negative. Earlier we determined the degree of homogeneity for several particular functions.
For instance, we found in Example 11.1.4 that the Cobb–Douglas functionF defined by F (x, y)=Axaybis homogeneous of degreea+b. Here is an even simpler example:
E X A M P L E 1 Show thatf (x, y)=3x2y−y3is homogeneous of degree 3.
Solution: If we replacexbyt xandybytyin the formula forf (x, y), we obtain f (t x, ty)=3(t x)2(ty)−(ty)3=3t2x2ty−t3y3=t3(3x2y−y3)=t3f (x, y) Thusf is homogeneous of degree 3. If we lett =2, then
f (2x,2y)=23f (x, y)=8f (x, y)
After doubling bothxandy, the value of this function increases by a factor of 8.
Note that the sum of the exponents in each term of the polynomial in Example 1 is equal to 3. In general, a polynomial is homogeneous of degreekif and only if the sum of the exponents in each term isk. Other types of polynomial with different sums of exponents in different terms, such asf (x, y)=1+xyorg(x, y)=x3+xy, are not homogeneous of any degree. (See Problem 6.)
Homogeneous functions of two variables have some important properties of interest to economists. The first isEuler’s theorem, which says that
f (x, y)is homogeneous of degreek ⇐⇒ xf1(x, y)+yf2(x, y)=kf (x, y) (2) It is easy to demonstrate that whenf is homogeneous of degree k, then the right-hand side of (2) is true. Indeed, differentiating each side of (1) w.r.t.t, using the chain rule to differentiate the left-hand side, gives
xf1(t x, ty)+yf2(t x, ty)=ktk−1f (x, y)
Puttingt =1 givesxf1(x, y)+yf2(x, y)=kf (x, y)immediately. Theorem 12.7.1 in the next section also proves the converse, and considers the case ofnvariables.
We note three other interesting general properties of functionsf (x, y)that are homo- geneous of degreek:
f1(x, y)andf2(x, y)are both homogeneous of degreek−1 (3) f (x, y)=xkf (1, y/x)=ykf (x/y,1) (forx >0,y >0) (4) x2f11(x, y)+2xyf12(x, y)+y2f22(x, y)=k(k−1)f (x, y) (5) To prove (3), keept andy constant and differentiate equation (1) partially w.r.t.x. Then tf1(t x, ty) = tkf1(x, y), so f1(t x, ty) = tk−1f1(x, y), thus showing thatf1(x, y) is homogeneous of degreek−1. The same argument shows thatf2(x, y)is homogeneous of degreek−1.
We can prove the two equalities in (4) by replacingtin (1) first by 1/xand then by 1/y, respectively.
Finally, to show (5) (assuming thatf (x, y)is twice continuously differentiable), we note first that becausef1(x, y)andf2(x, y)are both homogeneous of degreek−1, Euler’s theorem in (2) can be applied tof1andf2. It implies that
xf11(x, y)+yf12(x, y)=(k−1)f1(x, y) xf21(x, y)+yf22(x, y)=(k−1)f2(x, y)
(6) Let us now multiply the first of these equations byx, the second byy, and then add. Because f isC2, Young’s theorem implies thatf12 =f21, so the result is
x2f11(x, y)+2xyf12(x, y)+y2f22(x, y)=(k−1)[xf1(x, y)+yf2(x, y)]
By Euler’s theorem, however,xf1(x, y)+yf2(x, y)=kf (x, y). So (5) is verified.
E X A M P L E 2 Check properties (2) to (5) for the functionf (x, y)=3x2y−y3. Solution: We find thatf1(x, y)=6xyandf2(x, y)=3x2−3y2. Hence,
xf1(x, y)+yf2(x, y)=6x2y+3x2y−3y3 =3(3x2y−y3)=3f (x, y) Example 1 showed thatf is homogeneous of degree 3, so this confirms (2).
Obviouslyf1andf2are polynomials that are homogeneous of degree 2, which confirms (3). As for (4), in this case it takes the form
3x2y−y3=x3[3(y/x)−(y/x)3]=y3[3(x/y)2−1]
Finally, to show (5), first calculate the second-order partial derivatives, f11(x, y) = 6y, f12(x, y)=6x, andf22(x, y)= −6y. Hence,
x2f11(x, y)+2xyf12(x, y)+y2f22(x, y)=6x2y+12x2y−6y3=6(3x2y−y3)
=3ã2f (x, y) which confirms (5) as well.
S E C T I O N 1 2 . 6 / H O M O G E N E O U S F U N C T I O N S O F T W O V A R I A B L E S 433 E X A M P L E 3 Suppose that the production functionY =F (K, L)is homogeneous of degree 1. Show
that
Y /L=f (K/L), where f (K/L)=F (K/L,1)
(Thus, when a production function involving capitalK and labourL is homogeneous of degree 1, one can express the output–labour ratioY /Las a function of the capital–labour ratio K/L.) Find the form off whenF is the Cobb–Douglas functionAKaLbwitha+b=1.
Solution: BecauseF is homogeneous of degree 1,
Y =F (K, L)=F (L(K/L), Lã1)=LF (K/L,1)=Lf (K/L)
(Really, this is just a special case of (4).) WhenF (K, L) =AKaL1−a, thenf (K/L)= F (K/L,1)=A(K/L)a. Withk=K/L, we havef (k)=Aka.
Geometric Aspects of Homogeneous Functions
Homogeneous functions in two variables have some interesting geometric properties. Let f (x, y)be homogeneous of degreek. Consider a ray in thexy-plane from the origin(0,0) through the point(x0, y0)=(0,0). An arbitrary point on this ray is of the form(t x0, ty0) for some positive numbert. If we letf (x0, y0)=c, thenf (t x0, ty0)=tkf (x0, y0)=tkc.
Above any ray in thexy-plane through a point(x0, y0), the relevant portion of the graph off therefore consists of the curvez=tkc, wheret measures the distance along the ray from the origin, andc=f (x0, y0). A function that is homogeneous of degreekis therefore completely determined if its value is known at one point on each ray through the origin.
(See Fig 1.)
In particular, letk=1 so thatf (x, y)is homogeneous of degree 1. The curvez=tkc lying vertically above each relevant ray through the origin is then the straight linez=t c.
Because of this, it is often said thatthe graph of a homogeneous function of degree 1 is generated bystraight lines through the origin. Fig. 2 illustrates this.
(x0, y0) (tx0, t y0) z
y
x c
tkc
z ⫽ f(x, y) z
y
x
Figure 1 Figure 2 fis homogeneous of degree 1
We have seen how, for a functionf (x, y)of two variables, it is often convenient to consider its level curves in thexy-plane instead of its 3-dimensional graph. What can we say about the level curves of a homogeneous function? It turns out thatfor a homogeneous function, even if onlyone of its level curves isknown, then so are all its other level curves. To see
this, consider a function f (x, y)that is homogeneous of degreek, and letf (x, y) = c be one of its level curves, as illustrated in Fig. 3. We now explain how to construct the level curve through an arbitrary point Anot lying onf (x, y) = c: First, draw the ray through the origin and the pointA. This ray intersects the level curvef (x, y) = c at a pointDwith coordinates(x1, y1). The coordinates ofAwill then be of the form(t x1, ty1) for some value of t. (In the figure,t ≈ 1.7.) In order to construct a new point on the same level curve asA, draw a new ray through the origin. This ray intersects the original level curvef (x, y)=cat(x2, y2). Now use the value oft found earlier to determine the new pointBwith coordinates(t x2, ty2). This new pointB is on the same level curve asA becausef (t x2, ty2)=tkf (x2, y2)=tkc=tkf (x1, y1)=f (t x1, ty1). By repeating this construction for different rays through the origin that intersect the level curvef (x, y)=c, we can find as many points as we wish on the new level curvef (x, y)=f (t x1, ty1).
f(x, y) ⫽ c C
D A B
y
x (x2, y2)
(x1, y1) (tx2, t y2)
(tx1, t y1)
Figure 3 Level curves for a homogeneous function
The foregoing argument shows that a homogeneous functionf (x, y)is entirely deter- mined by any one of its level curves and by its degree of homogeneity. The shape of each level curve of a homogeneous function is often determined by specifying its elasticity of substitution, as defined in (12.5.2).
Another point worth noticing in connection with Fig. 3 is that the tangents to the level curves along each ray are parallel. We keep the assumption that f is homogeneous of degreek. If the level curve isf (x, y)=c, its slope is−f1(x, y)/f2(x, y). At the pointA in Fig. 3 the slope is
−f1(t x1, ty1)
f2(t x1, ty1) = −tk−1f1(x1, y1)
tk−1f2(x1, y1) = −f1(x1, y1)
f2(x1, y1) (∗) where we have used equation (3), expressing the fact that the partial derivatives off are homogeneous of degreek−1. The equalities in(∗)state that the two level curves throughA andDhave the same slopes at those points. It follows that, at every point along a ray from the origin the slope of the corresponding level curve will be the same. Stated differently, after removing the minus signs,(∗)shows that the marginal rate of substitution ofyforx is a homogeneous function of degree 0.
S E C T I O N 1 2 . 7 / H O M O G E N E O U S A N D H O M O T H E T I C F U N C T I O N S 435
P R O B L E M S F O R S E C T I O N 1 2 . 6
1. Show thatf (x, y)=x4+x2y2is homogeneous of degree 4 by using (1).
2. Find the degree of homogeneity ofx(p, r)=Ap−1.5r2.08.
⊂SM⊃3. Show thatf (x, y)=xy2+x3is homogeneous of degree 3. Verify that the four properties (2) to (5) all hold.
4. See whether the functionf (x, y)=xy/(x2+y2)is homogeneous, and check Euler’s theorem.
5. Prove the CES functionF (K, L) = A(aK− +bL−)−1/is homogeneous of degree one.
ExpressF (K, L)/Las a function ofk=K/L. (See Example 3.)
6. Show thatf (x, y)=x3+xyis not homogeneous of any degree. (Hint:Letx=y=1. Apply (1) witht=2 andt=4 to get a contradiction.)
7. Use the equations (6) withk=1 to show that iff (x, y)is homogeneous of degree 1 forx >0 andy >0, thenf11(x, y)f22(x, y)−[f12(x, y)]2≡0.
8. Iff (x, y)is homogeneous of degree 2 withf1(2,3)=4 andf2(4,6)=12, findf (6,9).
HARDER PROBLEM
⊂SM⊃9. Prove that ifF (x, y)is homogeneous of degree 1, then the elasticity of substitution can be expressed asσyx = F1F2/F F12. (Hint: Use Euler’s theorem (2), together with (6) and the result in Problem 12.5.3.)