5 Appendix
This appendix shows how to derive equation (30). First, recall equations (23) and
(24). Given the level of B, how do и and a affect Ah7
dAh
du
dAh
da
B=const
F(Ah,Kh) - aKh
Fa (Ah, Kh) (1 - u) - Fa (Ah, Kh*) (1 - t)
-uK h
B=const
Fah (Ah, Kh) (1 - u) - Fah (Ah, Kh*) (1 - t) > 0
Note that
dAh
da
dAh / uKh
du ∖F (Ah,Kh)
aK h
(32)
Given the level of B, how do u and a affect Az?
dA
du
dA
da
B=const
B=const
F (Al,Kl) - aKl
Fa (A,Kl)(1 - u)
-uKl
F (A,Kl)(1 - u)
With (32) it follows for equation (28):
dw h'-u'>B' £(F-k>dAdBh'£(u(F'-k`da)dB
Solve for H'
<g ,-u '> U' tf‘ l<f-K>dAdB
JsC+ ((F h-κh) ^c^∙f dB
and replace H' in equation (29):
∂W
da
B++ rAh
(H' - U') u uKdAdB
Jb- A a
(H' - U') J+B+ JAAh [(F - K)] dAdB
B++ ∂Ah
,B u f h - kft) -dAdB
fB+ U (Fh - Kh) tA AA-} dB
JB ∖ V t da uKft /
Rearrange and write:
16
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