Question

For each of the following situations, find the utility maximizing basket.

a) I=60 , px = 1, py = 9, u(x,y) = 7x^(1/4)y^(3/4)

b) I = 40, px = 8, py = 4, u(x,y) = 8x^2 + 3y^2

c) I = 100, px = 4, py = 5, u(x,y) = 2x+5y

d) I = 9, px = 18, py = 3, u(x,y) = 5xy-15x

e) I = 60, px = 1, py = 2, u(x,y) = 4x^(1/2)+2y

Answer 1

In each case, Budget constraint is: I = x.px + y.py

When x and y are imperfect substitutes, Utility is maximized when MRS = MUx/MUy = px/py, where

MUx = u/x

MUy = u/y

(a) Budget constraint: 60 = x + 9y

MUx = 7 x (y/x)^3/4

MUy = 7 x (x/y)^1/4

MRS = y/x = 1/9

x = 9y

Substituting in budget constraint,

60 = x + x = 2x

x = 60/2 = 30

y = x/9 = 30/9 = 10/3

(b) Budget constraint: 40 = 8x + 4y, or 10 = 2x + y (dividing by 4)

MUx = 16x

MUy = 6y

MRS = 16x/6y = 8x/3y = 2/1 = 2

8x = 6y

y = 4x/3

Substituting in budget constraint,

10 = 2x + (4x/3) = 10x/3

x = 30/10 = 3

y = (4 x 3)/3 = 4

(c) Budget constraint: 100 = 4x + 5y

This is a linear utility function with linear indifference curve which touches both axes. Optimal bundle is at one of the corner points of indifference curve.

From budget constraint,

When x = 0, y = 100/5 = 20, therefore u = 2 x 0 + 5 x 20 = 100

When y = 0, x = 100/4 = 25, therefore u = 2 x 25 + 5 x 0 = 50

Since utility is higher when x = 0 and y = 20, this is the optimal bundle.

(d) Budget constraint: 9 = 18x + 3y, or 3 = 6x + y (dividing by 3)

MUx = 5y - 15

MUy = 5x

MRS = (5y - 15) / 5x = (y - 3) / x = 18/3 = 6

y - 3 = 6x

y = 6x + 3

Substituting in budget constraint,

3 = 6x + 6x + 3

12x = 0

x = 0

y = 6 x 0 + 3 = 3

NOTE: As per Answering Policy, 1st 4 parts are answered.