Hence, substituting for dp from (13), we have
dv
dp
-vmx
S - xp
S — (x — x)xm — xp
< 0.
Similarly, assuming p > p and vm > 0, it is found that
dv
dx
vm
dp
, (x
dx
— x) + (p — p)
= vm(p — p)
S — xp
S — (x — x)xm — xp
> 0.
Proposition 5
Differentiating (17) w.r.t. p yields
dP x + χ p
dp x1 x1
p + (1
x) dp
x17 j dpi
(25)
dP
dx)
a
x1
— —
(p — p) + [α-p—p ⅛ + (1
x1 p p
x d j dp
x1 dx)
(26)
Substituting for dp from (13) and using x = xs > x1, we obtain
dP αXx £aS1 + (1 — α)S2 — (1 — α)x2xm — xp — ax((xs — x1)]
dpi α[S1 — x1m(x1 — x)] + (1 — α)(S2 — x2x‰) — xp
Similarly, substituting for dp from (14), we have
29
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