106920 REPARACION ESCUELA MARCO AURELIO SOTO
F012059 DEMOLICION DE TECHO ESTRUCTURA CANALETA/PERLINES
Suppose that the function g defined on the convex subset U of Rn is quasiconcave and differentiable, and that the derivatives of g at x∗ are not all zero. Suppose also that the functions h1, . . . , hm defined on U are quasiconvex and differentiable. Then
the Kuhn-Tucker conditions are sufficient for x∗ to solve the problem of maximising g(x) on U subject to hj(x) ≤ kj for j = 1, . . . , m. The conditions are:
i. The first order conditions are satisfied at x∗, that is there is a vector λ ∈ Rm
such that the partial derivatives of the Lagrangian:
L(k∗, λ, x) = g(x) + λ[k∗− g(x)] = g(x) + m X j=1 λj[k∗j − hj(x)]
at x∗ are zero, so:
∂g(x∗) ∂xi − m X j=1 λj ∂hj(x∗) ∂xi = 0 for i = 1, 2, . . . , n or in vector notation: ∂L(k∗, λ, x) ∂x = Dg(x ∗ ) − λDh(x∗) = 0.
ii. The multipliers are non-negative, that is λj ≥ 0 for j = 1, . . . , m, or in vector
notation λ ≥ 0.
iii. The vector x∗ is feasible, that is x∗ ∈ U and hj(x∗) ≤ kj∗ for j = 1, . . . , m or in
vector notation h(x∗) ≤ k∗.
iv. The complementary slackness conditions are satisfied, that is λj[kj∗− hj(x∗)] = 0
for j = 1, . . . , m or in vector notation λ[k∗− h(x∗)] = 0.
Theorem 19
Proof. Suppose that the theorem does not hold, that is there is a point x in U with g(x > g(x∗)) and hj(x) ≤ kj for j = 1, 2, . . . , m. The x∗ are not all zero; thus part 3 of
Theorem 18 implies that as g(x > g(x∗)):
n X i=1 (xi− x∗i) ∂g(x∗) ∂xi > 0. (5.6)
If hj(x∗) < hj(x∗) complementary slackness implies that λj = 0 so:
n X i=1 (xi− x∗i) λj ∂hj(x∗) ∂xi = 0.
5.9. Solutions to activities
This equality also holds if all the derivatives of hj at x∗ are zero. If hj(x∗) = kj, then
hj(x∗) = kj, so if the derivatives of hj at x∗ are not all zero the non-negativity of
multipliers and part 2 of Theorem 18 imply that:
n X i=1 (xi− x∗i) λj ∂hj(x∗) ∂xi ≤ 0. From the last two expressions:
n X i=1 (xi− x∗i) m X j=1 λj ∂hj(x∗) ∂xi ! ≤ 0. (5.7)
From the Kuhn-Tucker first order conditions: ∂g(x∗) ∂xi = m X j=1 λj ∂hj(x∗) ∂xi so: n X i=1 (xi− x∗i) ∂g(x∗) ∂xi = n X i=1 (xi− x∗i) m X j=1 λj ∂hj(x∗) ∂xi ! .
But (5.6) and (5.7) state that the left-hand side of this equation is strictly positive, whereas the left-hand side is not positive, which is impossible. As (5.6) and (5.7) follows from the assumption this contradiction implies that the result holds.
5.9
Solutions to activities
Solution to Activity 5.1From the definition if f is a quasiconvex function defined on the convex set U , then for all y in U the set on which f (x) ≤ f (y) is convex. But this is exactly the same as the set on which −f (x) ≥ −f (y), so this set is also convex for all y in U , which is what is required to make −f (x) a quasiconcave function.
Solution to Activity 5.2
A quasiconcave function defined on R could be increasing. In fact any increasing function defined on R is quasiconvex. This is because if f (x1) ≥ y, f (x2) ≥ y and
t ∈ [0, 1], then either x1 ≤ x2 in which case x1 ≤ tx1+ (1 − t)x2, so as f is increasing
f (y) ≤ f (x1) ≤ f (tx1+ (1 − t)x2), or x1 > x2 in which case as f is increasing
f (y) ≤ f (x2) ≤ f (tx1+ (1 − t)x2). In either case f (x1) ≥ y and f (x2) ≥ y implies that
f (tx1+ (1 − t)x2) ≥ y which is what is needed to show that f is quasiconvex.
Solution to Activity 5.3
A quasiconvex function cannot increase and then decrease as shown in Figure 5.1, because the set on which f (x) ≤ f (y1) includes y1 and y2, but not the points between y1
and y2, which have the form ty1+ (1 − t)y2 with t ∈ [0, 1]. Hence the set on which
Figure 5.4: The graph of f (x) = exp −12x2.
Solution to Activity 5.4
For the function f (x) = exp −12x2 the first derivative is −x exp −12x2 and the second derivative is (x2− 1) exp −1
2x
2. The graph is shown in Figure 5.4.
Solution to Activity 5.5
(a) The second derivative of exp −12x2 is positive when x < −1 or x > 1 and negative
when −1 < x < 1, so the function is neither concave nor convex on R, that is the entire real line.
(b) The function is quasiconcave because the set on which f (x) ≥ f (y) is the set of points with −y ≤ x ≤ y if y > 0, and the set on which y ≤ x ≤ −y if y < 0, and these sets are convex.
(c) The function is not quasiconvex, because for example the set on which f (x) ≤ f (2) is the set of points with either x ≤ −2 or x ≥ 2 which is not convex.
Solution to Activity 5.6
(a) The second derivative of exp −12x2 is positive when x > 1 and negative when
0 < x < 1, so the function defined on [0, ∞) is neither concave nor convex.
(b) The function is quasiconcave on [0, ∞) because the subset of [0, ∞) on which f (x) ≥ f (y) is the set of points with 0 ≤ x ≤ y which is convex.
(c) The function is quasiconvex on [0, ∞) because the subset of [0, ∞) on whcich f (x) ≤ f (y) is the set of points with y ≤ x which is convex.
5.9. Solutions to activities
Figure 5.5: The circle on which x21 + x22 ≤ 25 and its tangent at (3, 4).
Solution to Activity 5.7
The set on which f (x1, x2) = x21+ x22 ≤ 5 is the set of points on or within the circle with
radius 5 centred on the origin in Figure 5.5. The set is convex, and the function f (x1, x2) is quasiconvex. The tangent line at the point (3, 4) has equation:
(x1− 3) ∂f (3, 4) ∂x1 + (x2 − 4) ∂f (3, 4) ∂x2 = 0.
As the partial derivatives:
∂f (3, 4) ∂x1
= 6, ∂f (3, 4) ∂x2
= 8
this line has equation:
(x1− 3)6 + (x2− 4)8 = 0
or equivalently:
6x1+ 8x2 = 50.
The line is shown in Figure 5.5.
There are no points on this line with x2
1+ x22 < 25, but the point (3, 4) with
x21+ x22 = 25 lies on this line. The point (3, 4) solves the problem of maximising 6x1+ 8x2 subject to x21+ x22 ≤ 25.
A reminder of your learning outcomes
By the end of this chapter, you should be able to:
state the formal definitions and explain the implications of quasiconcave and quasiconvex functions
explain the relationship between quasiconcave and concave functions
explain the relationship between quasiconvex and convex functions
explain the relationship between concavity and returns to scale for production functions
explain the relationship between the convexity assumption in consumer theory and quasiconcavity of the utility function
explain the relationship between the sets on which a quasiconcave function f (x) ≥ f (y) and the tangent line
prove the sufficiency of the Kuhn-Tucker conditions in case of a quasiconcave objective function and quasiconvex constraints
use the Kuhn-Tucker conditions to solve optimisation problems with a quasiconcave objective function and quasiconvex constraints.
5.10
Sample examination questions
1. Find a solution to the consumer’s problem of maximising u(x) = xα1xβ2 on R2+ subject to p1x1+ p2x2 ≤ m. Assume that α > 0 and β > 0 but do not assume that
α + β = 1.
2. (a) A firm wants to find the combination of inputs K (capital) and L (labour) that minimise the cost rK + wL of producing output q subject to the production constraint K3L2 ≥ q. Assume that w = r = 1 and solve the problem. Find also the minimum cost of producing q.
(b) A firm wants to find the combination of inputs K (capital) and L (labour) that minimise the cost rK + wL of producing output q subject to the production constraint K3L2 ≥ q and a constraint on the supply of capital
K ≤ ¯K. As before assume that w = r = 1 and solve the problem. Find also the minimum cost of producing q.
5.11
Comments on the sample examination questions
1. The solution is:
x1 = α α + β m p1 x2 = β α + β m p2 .
5.11. Comments on the sample examination questions
You solve the problem using exactly the same Lagrangian techniques given in Chapter 2 for the solution of the consumer’s problem with the utility function u(x1, x2) = xa1xb2 where a > 0, b > 0 and a + b = 1. However the theorems on
optimisation in Chapter 2 work with a concave objective function and convex constraint functions. If α + β > 1 the objective function is quasiconcave but not concave. So you need to appeal to theorems on optimisation with quasiconcave and quasiconvex functions.
You need to state that the objective function is quasiconcave and explain why. (This is covered in Example 5.2). The constraint function is linear so concave and quasiconcave. The partial derivative ∂u/∂x1 = axa−11 xb2 > 0. Thus the Kuhn-Tucker
conditions are sufficient for a solution. You then find values of x1, x2 and the
multiplier that satisfy the Kuhn-Tucker conditions.
2. (a) As in Question 1 you need to name the theorem you are using, and check that the conditions hold. Here you are solving a constrained minimisation problem of minimising K + L with a ≥ constraint K3L2 ≥ q. You can turn this into a
constrained maximisation problem with a ≤ constraint by multiplying by −1 so the objective is −K − L and the constraint is −K3L2 ≤ −q. The objective is linear so concave and quasiconcave. From the previous question the function K3L2 is quasiconcave, so −K3L2 is quasiconvex. Hence this is a problem of
maximising a quasiconcave function subject to a quasiconvex constraint. The partial derivatives of the objective are not zero. Thus the Kuhn-Tucker conditions are sufficient for a solution. The Lagrangian is:
L = −K − L + λ(−1 + K3L2). The first order conditions are:
∂L ∂K = −1 + 3λK 2L2 = 0 ∂L ∂L = −1 + 2λK 3L = 0 so: 3λK2L2 = 1 2λK3L = 1
which implies that:
2K 3L = 1
so K = (3/2)L. The first order conditions cannot be satisfied if λ = 0, so λ > 0 and complementary slackness implies that the constraint K3L2 ≥ q must be
satisfied as an equality so q = K3L2. As K = (3/2)L this implies that:
L = 2 3 3/5 q1/5 and as K = (3/2)L: K = 3 2 2/5 q1/5.
These values of K and L are feasible and satisfy the first order conditions. From the first order conditions the multiplier λ = (1/2)K−3L−1 > 0 so the multiplier is nonnegative. Complementary slackness is satisfied because the constraint is satisfied as an equality. Thus the Kuhn-Tucker conditions are satisfied so this is a solution to the problem. The minimum cost of production is: K + L = 2 3 3/5 + 3 2 2/5! q1/5.
(b) This is the same problem as before except for the additional constraint that K ≤ ¯K. This constraint is linear so quasiconvex. From the argument in the previous question all the conditions of the Kuhn-Tucker theorem with quasiconcave objective and quasiconvex constraints are satisfied. The Lagrangian is:
L = −K − L + λ(−q + K3L2) + µ( ¯K − K).
If µ = 0 this is the same as the Lagrangian for the previous question, and :
K = 3 2 2/5 q1/5 L = 2 3 3/5 q1/5
solve the Kuhn-Tucker conditions for the new problem provided K is feasible, which requires that K ≤ ¯K, so:
q ≤ 2 3
2 ¯ K5.
The complementary slackness condition on the constraint K ≤ ¯K is satisfied because µ = 0.
However if q > (2/3)2K¯5 this solution is not feasible. Looking for a solution with K = ¯K, the constraint K3L2 ≥ q is satisfied as an equality if:
L = q1/2K¯−3/2.
The first order condition for L is:
−1 + 2λK3L = 0 which implies that:
λ = 1 2K3L = 1 2 ¯ K−3/2q−1/2
so the multiplier λ is non-negative. The first order condition for K is:
−1 + 3λK2L2− µ = 0 so: µ = 3λK2L2− 1 = 3 2K¯ −5/2 q1/2− 1.
5.11. Comments on the sample examination questions
The multiplier µ is this non-negative if:
q ≥ 2 3
2 ¯ K5.
Complementary slackness is satisfied because the constraints K3L2 ≥ q and
K ≤ ¯K are satisfied as equalities. The minimum cost of production is:
K + L = ¯K + q1/2K¯−3/2.
Summing up:
◦ If q ≤ 232K¯5 the solution is K = 322/5q1/5, L = 233/5q1/5. The minimum cost of production is 233/5+ 322/5q1/5.
◦ If q > 232K¯5 the solution is K = ¯K, L = q1/2K¯−3/2. The minimum cost
Chapter 6
Expenditure and cost minimisation
problems
6.1
Aim of the chapter
This chapter introduces the consumer’s expenditure minimisation problem, and derives the properties of its solution (compensated demand) and minimum value function (the expenditure function). It shows that the firm’s cost minimisaton problem is
mathematically the same as the consumer’s expenditure minimisation problem. The solution to the firm’s cost minimisation problem (conditional factor demand) and the minimum cost function (the cost function), have the same mathematical properties as compensated demand and the expenditure function. Finally, the chapter explores the close relationship between the solutions to the consumer’s utility maximisation and expenditure minimisation problems, and shows that Roy’s identity and the Slutsky equation derive from this relationship.
6.2
Learning outcomes
By the end of this chapter, you should be able to:
define compensated demand
describe the relationship between compensated demand, uncompensated demand, income and substitution effects
outline the properties of compensated demand
state the definition and properties of the expenditure function
explain the relationship between the firm’s cost minimisation problem and the consumer’s expenditure minimisation problem
explain the relationship between the solutions to the consumer’s utility maximisation problem and expenditure minimisation problem
describe the relationship between uncompensated demand, the indirect utility function, compensated demand and the expenditure function
state and derive Roy’s identity
Figure 6.1:Income and substitution effects.
6.3
Essential reading
This chapter is self-contained and therefore there is no essential reading assigned. But you may find further reading very helpful.
6.4
Further reading
Varian, H.R. Intermediate Microeconomics. For a reminder of the intermediate microeconomics treatment of consumer theory read Chapter 8 on the Slutsky equation, and Chapter 20 on cost minimisation.
Varian, H.R. Microeconomic Analysis. For a more sophisticated treatment
comparable to this chapter read Chapters 5, 6 and 9 or the relevant section of any mathematical treatment of the theory of the consumer and the firm.