Problem set 9, solutions

1. Take f : Z → R to be a production function associated with a single-output single-input technology Y where Z ⊆ R≥0. You know that Y asso-ciated with f is

Y = {(−z, q) ∈ −Z × R|q ≤ f(z)}.

For each property p of Y out of those discussed in the lecture (exceptnon-empty and additive), state a property p′ of f , such that if f satisfies p′,then the associated Y satisfies property p.1 For each property p′ you state,give a (possibly graphical) example of f that violates p′ and show that theassociated Y violates p.

solution: We need to find p′ for closed, no free lunch, possibility of in-action, free disposal, irreversibility, non-increasing returns to scale, non-decreasing returns to scale, constant returns to scale and convex propertiesof Y . Note also that the problem is asking for p′ that is a sufficient condi-tion for p (formally p′ ⇒ p). To see that necessary condition is not askedfor (necessary would mean p ⇒ p′, or, equivalently, ¬p′ ⇒ ¬p), the exam-ples that we are asked for are supposed to satisfy ¬p′ ⇒ ¬p, but for one(example) of ¬p′. Below, unless explicitly stated, assume that Z = R≥0.

Closed: Sufficient condition is that f is continuous. For any z ∈ R≥0 wehave (−z, q) ∈ Y if q ≤ f(z) so that any boundary of Y is when z = 0 orwhen q = f(z). Continuity is thus clearly sufficient.

Example of discontinuous f inducing Y that violates closedness is below.Notice also that if the value of f at the jump was qh, then Y would still beclosed (hence continuity is sufficient but not necessary, I put on record herea conjecture that sufficient and necessary condition is upper semicontinuityof f).

q

z

ql

qh

Y

No free lunch: Clearly f(0) ≤ 0 is sufficient (and necessary in this case).Hence an example that violates p′ and implies no free lunch violation is anyf(0) > 0.

Possibility of inaction: Sufficient (and again necessary in this case) con-dition is that f(0) ≥ 0. This in fact involves two conditions. First, that

1 For returns to scale and convexity you may assume f(0) = 0.

1

Problem set 9, solutions

f is defined at 0, i.e., 0 ∈ Z. Second, that f(0) ≥ 0. Hence one possible(but strong) sufficient condition is that Z = R≥0 and f(0) = 0. An exam-ple of violation of possibility of inaction is f(0) = −1. Another example isZ = [1,∞).

Free disposal: When f(z) is defined, any (−z, q) with q ≤ f(z) belongsto Y and hence output can be freely disposed of. This needs not be incase for input. Hence sufficient (and again necessary in this case) conditionis that f is non-decreasing. To see this more clearly, suppose f is non-decreasing. Suppose (−z, q) ∈ Y . This implies f(z) ≥ q. We need to makesure that any (−z′, q′) with z′ ≥ z and q′ ≤ q also (−z′, q′) ∈ Y . Becausef(z′) ≥ f(z) ≥ q ≥ q′ this is indeed the case. An example of non-monotonef is given below. It should be clear that the induced Y violates the freedisposal property.

q

z

y

Y

Irreversibility: This property is necessarily satisfied for any f at z ∈ Zsuch that z > 0. To see this note that if (−z, q) ∈ Y for some z > 0, thenz ∈ Z and −z /∈ Z since Z ⊆ R≥0. Hence sufficient (and necessary in thiscase) condition is f(0) ≤ 0 if z ∈ Z. To see this we know irreversibility holdsat any z ∈ Z such that z > 0. When 0 ∈ Z then the irreversibility holds ifand only if f(0) ≤ 0 for then (0, q) ∈ Y ⇔ q ≤ f(0) and hence (0,−q) /∈ Ysince −q ≥ f(0) (with equality only if q = f(0) = 0 but then the definitionof irreversibility has no bite). An example of violation of irreversibility isf(0) = 1, which implies (0, 12) ∈ Y as well as (0,−1

2) ∈ Y .Non-increasing returns to scale: Assuming f(0) = 0, any concave f will

clearly induce Y that satisfies the non-increasing returns to scale property.To see this more formally, take any (−z, q) ∈ Y . Then f(z) ≥ q. We need toprove (−αz, αq) ∈ Y for any α ∈ [0, 1], or f(αz) ≥ αq. By concavity, for anyα ∈ [0, 1] we know f(αz + (1− α)0) ≥ αf(z) + (1− α)f(0) or, equivalently,f(αz) ≥ αf(z). Since f(z) ≥ q, f(αz) ≥ αf(z) ≥ αq, as desired. Anexample of violation of non-increasing returns to scale is convex f .

Non-decreasing returns to scale: Assuming f(0) = 0, any convex f willclearly induce Y that satisfies the non-decreasing returns to scale property.To see this more formally, take any (−z, q) ∈ Y . Then f(z) ≥ q. We needto prove (−αz, αq) ∈ Y for any α ≥ 1, or f(αz) ≥ αq. Fix α > 1 and

2

Problem set 9, solutions

assume, towards a contradiction, that f(αz) < αq (for α = 1 the inequalityhas to fail). By convexity, f(βαz+ (1−β)0) ≤ βf(αz) + (1−β)f(0) for anyβ ∈ [0, 1], or, equivalently, f(βαz) ≤ βf(αz). Taking β = 1

α ∈ (0, 1), thisbecomes f(z) ≤ 1

αf(αz), and since f(αz) < αq, f(z) ≤ 1αf(αz) < 1

ααq = qand hence f(z) < q, contradicting f(z) ≥ q.

Constant returns to scale: Clearly any strictly convex f induces Y thatfails the non-increasing returns to scale property and any strictly concavef induces Y that fails the non-decreasing returns to scale property. Sinceconstant returns to scale is the union of the two earlier properties, it isintuitive that f has to be convex and concave, that is linear. Any non-linearf is going to violate constant returns to scale, hence linearity is sufficientand necessary condition.

Convex: Clearly any Y with increasing returns to scale is going to violateconvexity so that we have to rule out strictly concave f . Concave f isthus sufficient for convex Y . To see this more formally, pick (−z, q) ∈ Yand (−z′, q′) ∈ Y . Hence f(z) ≥ q and f(z′) ≥ q′. We need to show(−αz − (1− α)z′, αq + (1− α)q′) ∈ Y for any α ∈ [0, 1]. By concavity of f ,f(αz + (1− α)z′) ≥ αf(z) + (1− α)f(z′) for any α ∈ [0, 1]. Hence, for anyα ∈ [0, 1], f(αz + (1 − α)z′) ≥ αf(z) + (1 − α)f(z′) ≥ αq + (1 − α)q′, asdesired.

2. Give an example of a production set Y that satisfies the constant returnsto scale property but is not convex.

solution: The figure below is an example that the question asks for. Theclaimed properties (constant returns to scale but not convex) are easy tocheck.

q

zY

Y

3. Consider a production function f : R≥0 → R associated with single-input single-output technology. Suppose that f(0) = 0 and that the associ-ated Y satisfies the non-decreasing returns to scale property. Let p > 0 bethe market price of output and w > 0 the market price of input. Show thatif (PMP ) has a solution, then π(p, w) = 0.

3

Problem set 9, solutions

solution: (PMP ) for the current case writes as maxz≥0 pf(z)−wz. Sup-pose z∗ solves (PMP ) and π(p, w) = pf(z∗)− wz∗ = a > 0 (note that thisimplies z∗ > 0). By the non-decreasing returns to scale property, f(2z∗) ≥2f(z∗) and hence we have pf(2z∗)−w2z∗ ≥ 2pf(z∗)−2wz∗ = 2a > a, wherethe last inequality follows from a > 0. Thus 2z∗ > z∗ gives strictly highervalue of the objective function in (PMP ) than z∗ and thus z∗ cannot be asolution to (PMP ).

4. Let c(w, q) be the value function of (CMP ) (with concave productionfunction), that is, it is a cost function with a vector of input prices w andoutput q. Suppose c(w, q) is differentiable in q. Show that marginal costsequal average costs at the level of output that minimizes the latter.

solution: Hold w fixed throughout. Note that since the production func-tion is concave, c(w, q) will be convex in q. Minimizing ac(q) = c(w,q)

q is

associated with necessary and sufficient condition ac(q)′ = 0 and ac(q)′′ ≥ 0.

ac(q)′ =c′(w, q)q − c(w, q)

q2

ac(q)′′ =c′′(w, q)q3 + c′(w, q)q2 − c′(w, q)q2 + (c′(w, q)q − c(w, q))2q

q4

implies that ac(q)′ = 0 is equivalent to c′(w, q) = ac(q). To see that ac(q)′′ ≥0, ac(q)′′ = c′′(w,q)

q when c′(w, q) = ac(q). Since c(w, q) is convex, ac(q)′′ ≥ 0.

5. Suppose a firm has a cost function c(w, q) over vector of input pricesw ≥ 0 and level of output q ≥ 0. Suppose that the market price of the firm’soutput is p > 0, that inaction is possible c(w, 0) = 0 and that c(w, q) isdifferentiable in q. Denote by AC(q) and MC(q) the average and marginalcosts of the firm at product level q respectively (holding w fixed). SupposeMC(q) < AC(q) for g ≥ 0. Show that the profit maximizing level of outputis zero.

solution: (PMP ) for the current case is maxq≥0 pq − c(w, q). Note thatsince inaction is possible, the value of the objective function in (PMP ) atq = 0 is pq − c(w, q) = 0. Hence the value of the objective function in(PMP ) evaluated at the problem’s solution, q∗, has to be non-negative.Necessary condition for profit maximization is p = c′(w, q∗) when q∗ > 0.Hence the objective function evaluated at q∗ > 0 is c′(w, q∗)q∗ − c(w, q∗) =q∗[MC(q∗) − AC(q∗)] < 0. Thus q∗ > 0 cannot be a solution to (PMP ),leaving only q∗ = 0 and we know that the objective function at q∗ = 0 iszero.

4

Problem set 9, solutions

6. Consider a production set Y in R2 given by

Y = {(y1, y2) ∈ R2|(y1 + y1)2 + (y2 + y2)

2 ≤ 1}

where y1 < −1 and y2 > 1 are parameters. Which of the production setproperties discussed in the lecture does Y satisfies and which it does not?

solution: Y given in the question is drawn below. Note that y1 < 0 andy2 > 0 for any (y1, y2) ∈ Y .

y2

y1y1

y2

Y

The properties Y satisfies clearly are: non-empty, closed, no free lunch,irreversibility, convexity. The properties Y clearly violates are: possibilityof inaction, free disposal, non-increasing returns to scale, non-decreasingreturns to scale, constant returns to scale, additivity.

7. Suppose the indirect utility of consumer i admits Gorman form repre-sentation. That is, given an L vector of commodity prices p and income wiof the consumer, it can be written as

vi(p, wi) = ai(p) + b(p)wi

where ai(p) and b(p) are functions of p only. Show that the income effect indemand for commodity l ∈ {1, . . . , L} is independent of wi (you may assumeenough differentiability you may need).

solution: From Roy’s identity Walrasian demand for good j by consumeri is

xij(p, wi) = −∂vi(p,wi)∂pj

∂vi(p,wi)∂wi

=

∂ai(p)∂pj

+ ∂b(p)∂pj

wi

b(p).

Hence∂xij(p,wi)

∂wi=

∂b(p)∂pj

b(p) , which is independent of wi.

8. Suppose a firm with single-output single-input technology has produc-tion function f : R≥0 → R of the form f(z) = az where a > 0 is constant.Suppose the market price of the firm’s output is p > 0 and the market priceof the firm’s input is w > 0.

5

Problem set 9, solutions

1. Solve firm’s (PMP ).

2. Solve firm’s (CMP ).

solution: 1. (PMP ) is maxz≥0 pf(z) − wz = maxz≥0 z(pa − w). Thesolution to the problem is z∗ where z∗ = 0 if pa − w < 0, z∗ = m wherem ∈ R≥0 if pa− w = 0 and does not exists if pa− w > 0.

solution: 2. (CMP ) is minz≥0wz subject to f(z) ≥ q. The constraintrewrites as az ≥ q, or, z ≥ q

a . If q ≤ 0 the solution is clearly z∗ = 0. Ifq > 0, the constraint will clearly bind in optimum so z∗ = q

a .

9. Suppose a firm with single-output technology has production functionf : R2

≥0 → R of the form f(k, l) = kαlβ where α > 0 and β > 0 are constants.Suppose the market price of the firm’s output is p > 0, the market price ofthe capital input k is wk > 0 and the market price of the labor input l is wl.

1. Write down the firm’s (PMP ).

2. Is the objective function of the firm’s (PMP ) convex? Is it concave?

3. Solve the firm’s (PMP ). Write down the profit function π(p, wk, kl)and the supply function y(p, wk, wl) associated with (PMP ). Confirmthe properties of those two functions discussed in the lecture.

4. Write down the firm’s (CMP ).

5. Is the objective function of the firm’s (CMP ) convex? Is it concave?

6. Write down the expression of the firm’s iso-production curve. Is theconstraint set in (CMP ) convex (heuristic argument suffices here)?

7. Write down the expression of the firm’s iso-cost curve.

8. Solve the firm’s (CMP ). Write down the cost function c(wk, wl, q)and the conditional factor demands k(wk, wl, q), l(wk, wl, q). Confirmthe properties of these function discussed in the lecture.

solution: 1. (PMP ) is maxk≥0,l≥0 pf(k, l)− wkk − wll.

solution: 2. Hessian of the objective function is[α(α− 1)kα−2lβ αβkα−1lβ−1

αβkα−1lβ−1 β(β − 1)kαlβ−2

]with determinant equal to

αβ[1− α− β]k2(α−1)l2(β−1).

6

Problem set 9, solutions

When α + β < 1 the objective function is clearly strictly concave. Whenα+ β = 1 the objective function is concave. When α+ β > 1, the objectivefunction is neither concave nor convex.

solution: 3. The Lagrangian of the problem is

L(k, l) = kαlβ − wkk − wll − µk[−k]− µl[−l]

where µk and µl are multipliers associated with the non-negativity con-straints. The system of first order conditions is

pα(k∗)α−1(l∗)β − wk + µ∗k = 0

pβ(k∗)α(l∗)β−1 − wl + µ∗l = 0

if (k∗, l∗) solves (PMP ). In general there are four cases to consider depend-ing on µ∗k ∈ {0,R>0} and µ∗l ∈ {0,R>0}. Let us dispose of the two easycases: µ∗k = 0 and µ∗l > 0 or µ∗k > 0 and µ∗l = 0. In the former case l∗ = 0,which means that the first first order condition cannot be satisfied. In thelatter case k∗ = 0, which means that the second first order condition cannotbe satisfied.

The first case that can indeed arise is µ∗k > 0 and µ∗l > 0. This impliesk∗ = l∗ = 0 (and µ∗k = wk > 0, µ∗k = wl > 0), in which case the value of theobjective function is zero.

The second case that can indeed arise if µ∗k = µ∗l = 0. The foc systemcollapses to

pα(k∗)α−1(l∗)β − wk = 0

pβ(k∗)α(l∗)β−1 − wl = 0

and after some algebra solves to

k∗ =

[pββ

αβ−1wβ−1k

wβl

] 11−α−β

l∗ =

[pαα

βα−1wα−1l

wαk

] 11−α−β

.

(SC)

The problem is that this k∗ and l∗ need not solve the problem. We need togo through three cases now.

Case 1: α+ β > 1. We know that the objective function is not concaveso the solution to the collapsed foc system we derived should be suspect. Infact, along any ray parameterized by k = cl where c ∈ R>0, the collapsedsystem rewrites as

pα(c)α−1lα+β−1 − wk = 0

pβ(c)αlα+β−1 − wl = 0

7

Problem set 9, solutions

which makes it clear that when it holds for some l = l∗, it has to fail for anyl > l∗ with both of the derivatives being strictly positive.

This still leaves the possibility that k∗ = l∗ = 0 solve the (PMP ). To seethat this cannot be the case (and hence to see that the solution to (PMP )does not exists when α + β > 1) note that along any ray parameterized byk = lc with c ∈ R>0 the objective function of (PMP ) is lα+βpcα−l(wkc+wl).Second derivative of this function w.r.t. l is (α+ β)(α+ β− 1)pcαlα+β−2 sothat it is convex. Hence its limit as l→∞ is ∞.

Case 2: α + β = 1. The k∗ and l∗ derived above are clearly undefinedwhen α + β = 1. There are several ways how to proceed. One is to find amaximal output on an iso-cost line. This will pin down optimal proportionof inputs the firm wants to use (note that α+β = 1 implies that on any rayk = cl the output is linear function of l).

Any iso-cost line is given by kwk + lwl = c for some c ∈ R, or k = c−lwlwk

.Maximal output on this line is l that solves

∂

∂l

[p

(c− lwlwk

)αlβ]

= 0

which after some algebra yields

lαwlβwk

=c− wllwk

= k.

One can check that this is indeed maximum: for any k = 0 or l = 0 the out-put is zero, it is strictly positive in the interior and we know the productionfunction, which has the same Hessian as the objective function of (PMP ),is concave when α+ β = 1.

Substituting k = l αwlβwkinto the objective function of (PMP ) gives

p

(αwlβwk

)αlαlβ −

(wk

αwlβwk

l + wll

)= l

[p

(αwlβwk

)α− wlβ

].

The term in the square brackets is zero if p =(wlβ

)β (wkα

)α. Let me denote

p =(wlβ

)β (wkα

)α. If p > p then (PMP ) has no solution as the objective

function is increasing in l. When p = p, then (PMP ) has infinite numberof solutions l∗ = a, k∗ = αwl

βwka where a ∈ R≥0. When p < p, then (PMP )

has unique solution l∗ = 0 and k∗ = 0.Case 3: α + β < 1. In this case the solution k∗ and l∗ to the collapsed

system of foc’s derived above and given in (SC) is well defined. We still donot know if it constitutes a solution to (PMP ), as it can be the case thatthe objective function evaluated at the k∗ and l∗ is non-positive, in whichcase the solution to (PMP ) would be no production.

8

Problem set 9, solutions

To determine which case applies we can use the similar optimum on theiso-cost curve technique from above. The derivation of k = l αwlβwk

is identical.The objective function of (PMP ) is

p

(αwlβwk

)αlα+β − lwl(α+ β)

β.

Now notice that the first derivative of this function w.r.t. l is

(α+ β)p

(αwlβwk

)αlα+β−1 − wl

α+ β

β

which equals zero at l∗ stated in (SC) but, most importantly, is positiveas l → 0. To see this, note that lα+β−1 = 1

l1−α−β. Since 1 − α − β > 0,

liml→0 l1−α−β = 0, which means that the first term goes to ∞ as l → 0.

Hence, along the k = l αwlβwkray, the objective function in (PMP ) is increasing

for small l, which means that (0, 0) cannot solve (PMP ). Hence l∗ and k∗

given in (SC) constitute the solution to (PMP ).To summarize, if α+β > 1 there is no solution to (PMP ). If α+β = 1,

either there is no solution, or there is a solution in which the profit functionis zero. If α+ β < 1, then

k(p, wk, wl) =

[pββ

αβ−1wβ−1k

wβl

] 11−α−β

l(p, wk, wl) =

[pαα

βα−1wα−1l

wαk

] 11−α−β

.

are factor demands in (PMP ). These can be used to find the supply functionand the profit function. Some algebra gives

y(p, wk, wl) =

[pα+β

ααββ

wαkwβl

] 11−α−β

π(p, wk, wl) = py(p, wk, wl)− wkk(p, wk, wl)− wll(p, wk, wl).

Note that the notation here for the supply function includes only the output,but the factor demands should be part of the vector of supply function aswell.

To confirm the properties discussed in the lecture, we first note thaty(p, wk, wl), k(p, wk, wl) and l(p, wk, wl) are all homogeneous of degree zeroin (p, wk, wl). Hence π(p, wk, wl) is homogeneous of degree one in (p, wk, wl).

To confirm that

∇[p,wk,wl]Tπ(p, wk, wl) = [y(p, wk, wl), k(p, wk, wl), l(p, wk, wl)]

T

9

Problem set 9, solutions

we need to show that (you should see why)

p∂y(p, wk, wl)

∂p− wk

∂k(p, wk, wl)

∂p− wl

∂l(p, wk, wl)

∂p= 0

p∂y(p, wk, wl)

∂wk− wk

∂k(p, wk, wl)

∂wk− wl

∂l(p, wk, wl)

∂wk= 0

p∂y(p, wk, wl)

∂wl− wk

∂k(p, wk, wl)

∂wl− wl

∂l(p, wk, wl)

∂wl= 0.

This is potentially a lot of algebra. On the other hand notice that ∂k(z)αl(z)β

∂z =αk(z)α−1l(z)β + βk(z)αl(z)β−1. Simple substitution also shows

k(p, wk, wl)α−1l(p, wk, wl)

β =wkαp

k(p, wk, wl)αl(p, wk, wl)

β−1 =wlβp

With this we have

p∂y(p, wk, wl)

∂p− wk

∂k(p, wk, wl)

∂p− wl

∂l(p, wk, wl)

∂p=

p

[αk(p, wk, wl)

α−1l(p, wk, wl)β ∂k(p, wk, wl)

∂p

]+

p

[βk(p, wk, wl)

αl(p, wk, wl)β−1∂l(p, wk, wl)

∂p

]+

−wk∂k(p, wk, wl)

∂p− wl

∂l(p, wk, wl)

∂p=

∂k(p, wk, wl)

∂p

[pαwkαp− wk

]+∂l(p, wk, wl)

∂p

[βpwlβp− wl

]= 0

Similar derivation confirms the other two derivatives, w.r.t. to wk and wl,are zero as well. Convexity of π(p, wk, wl) is algebraically very demandingand will not be given here.

solution: 4. (CMP ) is mink≥0,l≥0wkk + wll subject to kαlβ ≥ q.

solution: 5. The objective function is concave and convex since it is linear.

solution: 6. The iso-production curve is given by f(k, l) = c for someconstant c. Standard techniques lead to the slope of this curve (in the spacewith k on the horizontal and l on the vertical axis) being − αl

βk . Drawing a

representative iso-production curve shows that the set (k, l) ∈ R2≥0 f(k, l) ≥

q is most likely convex.

solution: 7. The iso-cost curve has been derived above and is wkk+wll = cfor some constant c. Its slope (in the same space as above) is −wk

wl.

10

Problem set 9, solutions

solution: 8. For q ≤ 0 the solution to (CMP ) is (0, 0) so consider q > 0.The Lagrangian of the problem is

L(k, l, λ) = wkk + wll − λ[−kαlβ + q]

so that the first order conditions are

wk + λ∗α(k∗)α−1(l∗)β = 0

wl + λ∗β(k∗)α(l∗)β−1 = 0.

Some algebra gives

k(wk, wl, q) = q1

α+β

[wlα

wkβ

] βα+β

l(wk, wl, q) = q1

α+β

[wkβ

wlα

] αα+β

and

c(wk, wl, q) = (qwαkwβl )

1α+β

[α+ β

αα

α+β ββ

α+β

]c(wk, wl, q) is clearly homogeneous of degree one in (wk, wl) and non-

decreasing in q. Ignoring positive constants, Hessian of c(wk, wl, q) is αα+β ( α

α+β − 1)wα

α+β−2

k wβ

α+β

lα

α+ββ

α+βwα

α+β−1

k wβ

α+β−1

l

αα+β

βα+βw

αα+β−1

k wβ

α+β−1

lβ

α+β ( βα+β − 1)w

αα+β

k wβ

α+β−2

l

which has zero determinant and hence c(wk, wl, q) is concave in (wk, wl).k(wk, wl, q) and l(wk, wl, q) are clearly homogeneous of degree zero in (wk, wl).∂c(wk,wl,q)

∂wk= k(wk, wl, q) and ∂c(wk,wl,q)

∂wl= l(wk, wl, q) are a matter of simple

algebra to confirm.f homogeneous of degree one implies that α+ β = 1. We need to check

that c(wk, wl, q), k(wk, wl, q) and l(wk, wl, q) are then homogeneous of degreeone in q, which clearly holds. f concave implies that α + β ≤ 1. We needto check that c(wk, wl, q) is convex in q, which clearly holds.

11