Esercitazioni Microeconomia per Manager

(ES) b≥a≥c≥d

P

A a, a

c, c

B b, d

d, b

Π₁(p) = apq + bp(1-q) + dq(1-p) + c(1-p)(1-q) =

= apq + bp - bpq + dq - dpq + c - cp -cq + cpq =

= p[aq + d - dq] + bq - c(1-q) + bq - c(1-q) =

= p[(q(a-d-b+c) + d) - bq + c +cq + c(1-q) =

- c(-)

Retta in p decrescente: p*=0

Π₂(q) = apq + bp(1-q) + dq(1-p) + c(1-p)(1-q) =

= c - cp - dq

= apq + bp - bpq + dq - dpq +c - cp - cq + cpq =

= q[ap - bp + d - dp - c + cp] + (bq + c (1-p) ) =

= q[p(a-b-d+c) + d - c] + bq + c(1-p) =

= c(-)

Retta in q decrescente: q*=0

(p=0; q=0)

π(p) =

2pq + 3p(1-q) + 4q(1-p) + 4(1-p)(1-q) =

= 2pq + 3p - 3pq + 4q - 4pq + 4 - p - q + pq =

= 4pq + 2 - p - 3q + 1 =

p[-4q + 2] + 3q + 1 =

Retta in p decrescente: p* = 0

π(q) =

2pq + 4p(1-q) + 3q(1-p) + 4(1-p)(1-q) =

= 2pq + 4p - 4pq + 3q - 3pq + 4 - p - q + pq =

= 4pq + 3p + 2q - 4 =

q[-4p + 2r] + 3p + 1 =

Retta in q decrescente: q* = 0

(P=0; Q=0)

E3

bσ>c

P

Δ̅̅b,0

B

b,̅b̅

S

e

D

c

σ,̅c̅

t_c,̅b̅*

π( P̅̅)= b̅σq+cp(1-q)+σ c(1-p)(1-q)=

=σpq+σpq+c σp+cq+σp = p[bq+σ ]=σ f+σ q +σ ( 1-q).

np

pq=0

P*=1

π₂ (q)=σ +σp(1-q)+bq(1-p)+σ(sp)(1-q)=

=σp+σpq+bq-bpq+σ -σpq+σpq =

1 ( ) p(c-b+c σ)+loc (1-p)=t op +σ (1-p)=

p(1-c-b+σ)b b) :

eddings = 0

p

1. _-c-bσe) t b≈e<0j;
2. p< e-b
3. c(b+c b)

R ->

1. dyž
2. p.B
3. p*q[0,1]
4. p <b 0=1

p

ansom: (p=1, q=0)

k

(f=1, q=0)

(3)

I: {10-x x} {----- -----} {-x x+y} II: {10- y² y} {------- -----} { 2 x+y} { - y² x}{------- -----}{ 2 x+y}

Π₁(x) = 10( x ) - x ( ---- ) ( x+y )

Π₂(y) = 10y( y ) - y² ( ---- ) ( x+y ) 2

∂Π₁ 10y ------ = ----- - 1 = 0; (x+y)² = 10y dx (x+y)²

∂Π₂ 10x ------- = ----- - y = 0 dy (x+y)²

d²Π₂ -2(10x) -20x -------- = ------ - 1 = 0 ; ------ = 1 ; (x+y)⁴ = -20x dy (x+y)⁴ (x+y)⁴

(4)

I: {25-x x} {----- -----} {-x x+y} II: {25 - 5y y} {------ -----} { -5y x+y} { -5y x} {------ -----} { 5 x+y}

Π₁(x) = 25( x ) - x ; ( --- ) ( x+y )

Π₂(y) = 25( y ) - 5y ; ( --- ) ( x+y )

∂Π₁ --------- = 25y dx (x+y)² - 4 = 0; (x+y)² = 25y

∂Π₂ --------- = 25x dy (x+y)² - 5 = 0; (x+y)² = 5x

25y = 5x ; 5y = x

3

U₁(b₁) = { v₁-b₂      b₁ ≥ b₂ -b₁      b₁ < b₂

U₂(b₂) = { v₂-b₁      b₂ > b₁ -b₂      b₂ ≤ b₁

• π₁ ∧ b₂ < v₁b₁: [b₂, ∞)
• π₁ ∧ b₂ = v₁b₁: {0} ∪ (v₁, ∞)
• π₁ ∧ b₂ > v₁b₁: ∅
• π₂ ∧ b₁ < v₂b₂: b₁ + ε
• π₂ ∧ b₁ = v₂b₂ = {0} ∪ (v₂, ∞)
• π₂ ∧ b₁ > v₂b₂ = 0

(b₂)(v₂)(b₁)(v₁)

π₁(x):

• x se y≥x
• 1-x se x≤y
• -x se x