a)
Given U=B0.4F0.6
Budget constraint is given by
5*B+10*F=100
or 100-5B-10F=0
So,
Maximize U=B0.4F0.6
Subject to
100-5B-10F=0
Lagrange is given by
L=B0.4F0.6+*(100-5B-10F)
b)
First order conditions give us
dL/dB=0.4B-0.6F0.6-5=0
or 0.4B-0.6F0.6=5 -----(1)
dL/dF=0.6B0.4F-0.4-10=0
or 0.6B0.4F-0.4=10 -----(2)
(100-5B-10F)=0 -----(3)
Divide equation (1) by equation (2), we get
(0.4B-0.6F0.6)/(0.6B0.4F-0.4)=5/10
(2/3)*(F/B)=1/2
or F=0.75*B
Set F=0.75B in equation 3, we get
100-5B-10*0.75B=0
100-12.5B=0
B=8 -----(optimal consumption of B)
F=0.75*B=0.75*8=6 -----(optimal consumption of F)
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