
Answer:
Explanation:
A point on U=800 is (5, 16)
From BL:
400*F+100D =4000
400*5+100*16 =3600<4000
Therefore u = 800 affordable.
U= 1200
F = 1200/10D
If D = 20
F = 1200/200
=6
Now from BL:
400*6+100*20= 2400+2000=4400>4000
Not affordable.
Maximization:
L = 10DF+ʎ[100*D+400*F – 4000]
Differentiating wrt D and F:
dL/dD = 10F + ÊŽ*100
dL/dF = 10D +ÊŽ*400
equating to zero; Â Â Â Â Â Â Â Â Â Â Â
ÊŽ= -F/10
ÊŽ=-D/40
equating the two:
F/10=D/40
D = 4F
From BL:
400*F+100*D = 4000
400F+100*4F = 4000
800F = 4000
F = 5
D = 4*5=20