1.learning Material-inventory 14.12.2018

MIT 14.04 Intermediate Microeconomie Theory Fall 2003

Answer keys for problem set 1 Hint If you don’t know how to answer a question, start by using what you know (definition, properties, etc.). It is always helpful if you can draw a graph and see the solution on it. Exercise 1.2 1 Consider the production function f (x) = (a1 xρ1 + a2 xρ2 ) ρ . In order to compute the elasticity of substitution, compute first the Technical Rate of Substitution T RS. Then take ln and differentiate w.r.t. ln xx21 . ∂f

T RS

=

1 =− − ∂x ∂f

a1 x1ρ−1

a2 x2ρ−1 a1 x2 ln + (1 − ρ) ln x1 a2 d ln xx21 1 = d ln |T RS| 1−ρ ∂x2

⇒ ln |T RS| = ⇒σ

=

Exercise 1.4 Use the definition of the elasticity of scale. Then find a relation between ∂f (tx) ∂f and ∂x (indeed f (tx) = f (tx1 , tx2 , ..., txn )). ∂t i n

t ∂f (tx) t � ∂f (tx) ∂ ln f (tx) = = xi ∂ ln t f (tx) ∂t f (tx) i=1 ∂xi n

⇒ (x) =

n

� xi ∂f (x) � ∂ ln f (tx) = i (x) |t=1 = ∂ ln t f (x) ∂xi i=1 i=1

Exercise 1.6 False. Check that f (x) = x3 is increasing and that f
(0) = 0, note that
f (x) > 0 for any other x. Exercise 1.8 Start with the definition of a homothetic function: f is homothetic then it exists a strictly monotonic function ϕ and a homogeneous function of degree 1 g such that: f (x1 , x2 ) = ϕ(g(x1 , x2 )) 1

Then compute the T RS at tx: ∂f

1 T RS(tx) = − ∂x ∂f

(tx)

∂x2 (tx)

∂g

1 = − ∂x ∂g

(tx)

∂x2 (tx)

Apply the following theorem to the function g: if h is a homogeneous function ∂h is a homogeneous function of degree k-1. of degree k, then ∂x i ∂g ∂g (tx) = (x) ∂xi ∂xi You can now conclude that T RS(tx) = T RS(x) Exercise 1.10 • let y and y
∈ Y

let t ∈ [0, 1]

Y is additive ⇒ ty ∈ Y and (1 − t)y
∈ Y

Y is additive ⇒ ty + (1 − t)y
∈ Y

then Y is convex

• let y ∈ Y prove by induction that ∀n ny ∈ Y

- n = 1 y ∈ Y

- assume ny ∈ Y

y and ny ∈ Y and Y additive ⇒ y + ny = (n + 1)y ∈ Y

• let y ∈ Y and t ≥ 0

call n such that n ≤ t < n + 1

previous result ⇒ ny ∈ Y

Y divisible (t − n)y ∈ Y

then by additivity ty ∈ Y ⇒ Y has CRtS

Exercise 2.2 Assume ∃x s.t. pf (x) − wx = π(x) > 0, let t > 1

IRtS ⇒ π(tx) = pf (tx) − wtx > ptf (x) − wtx = tπ(x)

π(tx)→∞ as t → ∞

2

Exercise 2.4 This question is about the elasticity of the ln of factor share Start by writting the factor share as a function of xx12 . ln ⇒

w2 x2 w1 x1

2 x2 ∂ ln w w1 x1

∂ ln xx12

= =

w2 x2 w1 x1

wrt to ln xx12 .

w2 x1 − ln w1 x2 2 ∂ ln w w1 −1 ∂ ln xx12

ln

At the profit maximizing bundle (y, x1 , x2 ), we have w1 = w2

∂f ∂x1 ∂f ∂x2

= |T RS|

⊆

Y ⊆YO

Hence the result. Exercise 2.6 By definition YI −

π (p) = π(p)

=

π + (p) =

max py

p∈Y I

max py p∈Y

max py

p∈Y O

As the set over which the maximization is performed becomes larger and larger, the profit cannot decrease: π − (p) ≤ π(p) ≤ π + (p) Exercise 3.2 Check first on a graph what the solution should look like. The profit line intersect is equal to πp and its slope is wp . When the profit line is flat enough (case 1 on the graph), the solution will be interior i.e. satisfying the FOC. When the profit line is steep (case 2 on the graph), it is optimal not to produce. The third profit lineon the graph is the one where the firm is indifferent between producing (y = 1, x = e) and not producing, both leading to zero profit. Let’s check: p max p ln x − wx ⇒ x = x w The profit should be non negative π ≥ 0 ⇒ wp ≥ e. If wp ≥ e, π(p, w) = p(ln wp − 1). If wp ≤ e, π(p, w) = 0. 3

Figure 1: Three profit lines using different sets of prices (p, w). In case 1, the profit maximizing bundle is an interior solution. In case2, the profit maximizing bundle corresponds to no production. The third line is the limit case where the firm is indifferent between producing or not. Exercise 3.4 xi (p, w)

=

y(p, w) =

π(p, w)

=

� � � � � � 1 a (1−aj ) 1−a1 −a2 aj j ai

p wj wi � � �a1 � �a2 ) � 1−a11−a2 a1 a2

a1 +a2 p w1 w2

� � �a1 � �a2 � 1−a 1−a 1 2 a1 a2 p (1 − a1 − a2 ) w2 w1

It must be the case that 0 < ai and a1 + a2 ≤ 1. Note that the previous functions are increasing in p and decreasing in wi .

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