Respuesta :
Answer:
Since F = 30 and C = 60, the optimal bundle of consumption is to consume 30 units of food and 60 units of clothing.
Explanation:
Maximize: [tex]U = F^{0.5} C^{0.5}[/tex]
Subject to: 120 = 2F + C
Setting up the Langrangian function, we have:
[tex]U = F^{0.5} C^{0.5}[/tex] + ∧(120 - 2F - C) .................... (1)
Where ∧ is used to represent lamda.
Obtain partial differentials of equation (1) with respect to F, C and ∧ as follows:
[tex]dU/dF = 0.5F^{-0.5} C^{0.5}[/tex] - 2∧ = 0 .................. (2)
[tex]dU/dC = 0.5F^{0.5} C^{-0.5}[/tex] - ∧ = 0.......................(3)
du/d∧= 120 - 2F - C = 0 ................................(4)
Divide equation (2) by (3) to eliminate ∧ and rearrange, we have:
C/F = 2
C = 2F ............................................................... (5)
Substitute C = 2F into equation (4), we have:
120 - 2F - 2F = 0
120 = 4F
F = 120/4
F = 30
Substitute F = 30 into equation (5), we have:
C = 2 * 30
C = 60
Since F = 30 and C = 60, the optimal bundle of consumption is to consume 30 units of food and 60 units of clothing. At this bundle, the consumer will maximize his utility.
