Question

Given the utility function:
U = X 3/4 . Y1/4
Find out the optimal quantities of the two commodities using Lagrange method and simplified procedure, if it is given that price of X is Rs.6 and price of Y is Rs.3 and income is equal to Rs.120.

Answer 1

Income = 120

Objective: Maximize U = $X^{0.75}*Y^{0.25}$

Subject to: Total Revenue (TR) = $P_X*X +P_Y*Y$

                       $120 = 6*X +3*Y$

L(λ, x, y) = $X^{0.75}*Y^{0.25} – λ(6X +3Y-120)$

$\frac{∂L}{∂X}$ = $0.75X^{-0.25}*Y^{0.25} - λ(6) = 0$

$\frac{∂L}{∂Y}$ = $0.25X^{0.75}*Y^{-0.75} – λ (3) = 0$

$\frac{∂L}{∂ λ}$ = -1(6X +3Y-120) = 0

Ratio of First order condition (FOC)

$\frac{∂L}{∂X} /\frac{∂L}{∂Y}$ = $\frac{λ6}{λ3} = 0.75X^{-0.25}*Y^{0.25} /0.25X^{0.75}*Y^{-0.75}$

$\frac{6}{3}$ = $\frac{0.75}{0.25}X^{-0.25-0.75}*Y^{0.25- -0.75}$

2 =$3X^{-1}*Y^{1}$

Y = $\frac{2}{3}X$

Solving for the equations:

6X +3Y= 120

$Y = \frac{2}{3}X$

$6X +3*\frac{2}{3}X = 120$

6X +2X = 120

8X = 120

X = $\frac{120}{8} = 15$

Solving for y

Y = $\frac{2}{3}X$

Y = $\frac{2}{3}*15$

Y = 10

Solving for Maximum Utility

U = $X^{0.75}*Y^{0.25}$

X = 15

Y = 10

U = $15^{0.75}*10^{0.25}$

U = $7.62199*1.77828$

U = 13.554