Answer to Question #273817 in Microeconomics for Jojo

Question #273817

Do each of a-d, both geometrically (you need not be precise) and using calculus. There are only two goods; x is the quantity of one good and y of the other. Your income is I and u(x,y) = xy + x + y.


(a) Px = $2; Py = $1; I = $15. Suppose Py rises to $2. By how much must I increase in order that you be as well off as before?


(b) In the case described in part (a), assuming that I does not change, what quantities of each good are consumed before and after the price change? How much of each change is a substitution effect? How much is an income effect?


(c) Px = $2; I =$15. Graph the amount of Y you consume as a function of Py , for values of Py ranging from $0 to $10 (your ordinary demand curve for Y).


(d) With both prices equal to $1, show how consumption of each good varies as I changes from $0 to $100.



1
Expert's answer
2021-12-07T12:47:22-0500

Solution

(a)

Maximize:

"U(x,y)=xy+x+y"

Subject to "I = P_x x + P_y y"

Lagrangian for the problem:

"L(x,y,\\lambda) = xy+x+y+\u03bb(I-P_xx-P_yy)"

First-order conditions(derivatives):


"\u2202L\/\u2202x=y+1-\u03bb^* P_x=0" (1)


"\u2202L\/\u2202y=x+1-\u03bb^* P_y=0" (2)


"\u2202L\/\u2202\u03bb=I-P_x x-P_y y=0" (3)


Using the first two conditions(1 and 2) we have

"(y+1)\/P_x=(x+1)\/P_y"

So,

"y^*=((P_x\/P_y)(x+1))-1\n\u200b" (4)


Substitute this into I function

"I=P_xx+P_y(((P_x\/P_y)(x+1))-1)"

"I=P_xx+P_xx+P_x-P_y"

"I=2P_xx+P_x-P_y" (5)

solve for x using (5)

"2P_xx=I-P_x+P_y"

"x^*=(I-P_x+P_y)\/2P_x" (6)

Substitute x* into (4)


"y^*=((P_x\/P_y)((I-P_x+P_y)\/2P_x+1))-1\n\u200b"


"y^*=(I-P_y+P_x)\/2P_y"



Demand function of Both Goods are:

"x^*=(I-P_x+P_y)\/2P_x"

"y^*=(I-P_y+P_x)\/2P_y"


I= 15, Px=2 and Py=1

Before Increase

"x^*=(15-2+1)\/2*2=3.5"

"y^*=(15-1+2)\/2*1=8"

Initial Utility

"U(3.5,8)=3.5(8)+3.5+8=39.5\\to40"

To make same utility after Py increased to $2

"x=(I+2-2)\/2*2=I\/4"

"y=(I+2-2)\/2*2=I\/4"

So,

"U(2,2)=(I\/4)^2+I\/4+I\/4=40"

"=(I^2+8I)\/16=40"

"I^2+8I-640"

Solving for I

"I=-8\u00b1 ( \u221a(64+(4*640)))\/2=(8\u00b151.22)\/2=\u00b121.61"


Income is non-negative so it's $21.61

Therefore I should be increased by: "21.61-15=" $6.61


(b)

After an increase of Py to $2

"x^*=(15-2+2)\/2*2=3.75"

"y^*=(15-2+2)\/2*2=3.75"

consumption of X and Y is 3.75 each

Substitution effect:

"Y= 8-3.75= 4.25"

"X=3.5-3.75=-0.25"

So more of X will be consumed and amount of Y consumed


(c)

Px = $2; I =$15

"Y=(I-P_y+P_x)\/2P_y =(15-P_y+2)\/2*2"

"Y=(17-P_y)\/4"


Graph of Y as a function of Py





(d)

"P_x=P_y=" $1

"x^*=(I-1+1)\/2(1)=I\/2"

"y^*=(I-1+1)\/2(1)=I\/2"


Therefore, when Income is at 0 the, demand for both x and y is 0, and as income increases from 0 to 100, their demand tends to increase also.

at 100 demands are:

"X=Y=100\/2=50"

So, demand for X, and Y become 50 each



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