Question #290904

Given utility function: U(x, y) = X1/3Y2/3, Px = 2, Py = 5 and M = 400, find:

a. The demand equation for X and Y.

b. The utility maximizing levels of X and Y.

c. The maximum utility.

d. The MRSx, y at the optimum level.


1
Expert's answer
2022-01-27T11:00:46-0500

U=(X13Y23)(X^ \frac {1}{3}Y^\frac{2}{3})

I= 400

x=2_x= 2

Py=5P_y= 5

a)Demand functions

For utility maxmization


13X23Y2323X13Y13=PxPy\frac{\frac{1}{3}X^\frac{-2}{3}Y^\frac{2}{3}}{\frac{2}{3}X^\frac{1}{3}Y^\frac{-1}{3}}=\frac {P_x}{P_y}


13Y23X=PxPy\frac{\frac {1}{3}Y}{\frac{2}{3}X}=\frac {P_x}{P_y}


23PxX=13PyY\frac{2}{3}P_xX=\frac{1}{3}P_yY


X= PyY2Px\frac{P_yY}{2P_x} 


2PxX=PyY2P_xX=P_yY


Y=2PxXPyY=\frac{2P_xX}{P_y} 

Plug into the budget constraint


m= PxX+PyYP_xX+P_yY

m=Px(PyY2Px)+PyYP_x(\frac{P_yY}{2P_x})+P_yY

2m=3PyY3P_yY


Y=2m3PyY^*= \frac{2m}{3P_y}

m=PxX+PyYP_xX+P_yY

m=PxX+PY(2PxXPy)m=P_xX+P_Y(\frac{2P_xX}{P_y})

m=PxX+2PxX=P_xX+2P_xX

X*=m3Px\frac{m}{3P_x}


b) Utility Maximizing levels

Y=2×4003×5=80015=53.33Y^*= \frac{2\times 400}{3\times 5}=\frac{800}{15}= 53.33


c) Maximum utility of consuming the two goods

= 66.3=120

d) MRS

MRS(xy)=MuxMuy=13X23Y2323X13Y13MRS_(xy)= \frac{Mu_x}{Mu_y}= \frac{\frac{1}{3}X^\frac{-2}{3}Y^\frac{2}{3}}{\frac{2}{3}X^\frac{1}{3}Y^\frac{-1}{3}}

=13Y23X=Y2X==\frac{\frac{1}{3}Y}{\frac{2}{3}X}=\frac{Y}{2X}=

=53.3366.67×2=53.33134.34=0.4=\frac{53.33}{66.67\times 2}=\frac{53.33}{134.34}=0.4


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Comments

Nkonde
02.10.23, 15:36

This was very helpful thank you

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