TD 5 - Corrigé - Marc Sangnier

C(q) = q2−4q+ 9
q
Cm(q) = ∂C(q)
∂q
Cm(q)=2q−4
CM(q) = C(q)
q
CM(q) = q2−4q+ 9
q=q−4 + 9 q
∂CM(q)
∂q = 0
⇐⇒ 1−9
q2= 0 ⇐⇒ 9 q2= 1 ⇐⇒ q2= 9
⇒q= 3
q= 3
CM(3) = 3 −4 + 9
3= 2
p > 2

Cm(q) = p
p > 2p < 2
p > 2
Cm(q) = p
⇐⇒ 2q−4 = p⇐⇒ 2q=p+ 4 ⇐⇒ q=p+ 4
2
s(p)
s(p) = 0 p < 2
s(p) = p+4
2p > 2
d(p) = 50
ppe
s(pe) = d(pe)
s(p) = d(p)
⇐⇒ p+ 4
2=50 p⇐⇒ p2+ 4p= 100 ⇐⇒ p2+ 4p−100 = 0
pe≈8,19
p= 2 d(2) = 25
q= 3 n
3n
25 = 3n
n=25
3
25
n
ε q p
ε=∂q
q/∂p p
ln {q(p1;p2;R)}=a.ln {p1}+b.ln {p2}+cR

∂q
q=a∂p1 p1
∂q
q=b∂p2 p2
∂q
q=c.∂R
εp1=a
εp2=b
εR=cR
t s(p) = ap d(p) = 10 −p
ape= 10 −pe⇐⇒ pe(a+ 1) = 10 ⇐⇒ pe=10
a+ 1
pt+1 −pt= 10 −pt−apt
pt+1 = 10 −apt
pt=pe
pe= 10 −ape
pt+1 −pe=−a(pt−pe)
a < 1