Question #72034

I. U=(x y)=x^(1/3)y^(2/3) that subjected to budget constant 6x+3y=720 then find
1. MUx,
2. MUy,
3. MRSx,y
4.write equilibrium optimum condition.
II. Assume u function of a consumers is u=x^(2) + y^(2) then, Required.
1. MRSy,x
III. a consumers u function give as u=x^(4). y^(-2)
1. MUx,
2. MUy,
3. MRSx,y
1

Expert's answer

2017-12-21T09:29:07-0500

Answer on question #72034, Economics / Microeconomics

I. U=(x y)=x^(1/3)y^(2/3) that subjected to budget constant 6x+3y=720 then find

1. Mux


Mux=U=(xy)=(1\3)x(2/3)y(2/3)\mathrm {M u x} = \mathrm {U} ^ {\prime} = (\mathrm {x y}) = (1 \backslash 3) \mathrm {x} ^ {\wedge} (- 2 / 3) \mathrm {y} ^ {\wedge} (2 / 3)


2. MUy


Muy=U=(xy)=(2\3)x(1/3)y(1/3)\mathrm {M u y} = \mathrm {U} ^ {\prime} = (\mathrm {x y}) = (2 \backslash 3) \mathrm {x} ^ {\wedge} (1 / 3) \mathrm {y} ^ {\wedge} (- 1 / 3)


3. MRSx,y


MRSxy=MUx/MUyM R S _ {x y} = M U _ {x} / M U _ {y}MRSx,y=(1\3)x(2/3)y(2/3):(2\3)x(1/3)y(1/3)=y/2x\mathrm {M R S x , y} = (1 \backslash 3) \mathrm {x} ^ {\wedge} (- 2 / 3) \mathrm {y} ^ {\wedge} (2 / 3): (2 \backslash 3) \mathrm {x} ^ {\wedge} (1 / 3) \mathrm {y} ^ {\wedge} (- 1 / 3) = \mathrm {y} / 2 \mathrm {x}


4. write equilibrium optimum condition:


(1\3)x(2/3)y(2/3)/6=(2\3)x(1/3)y(1/3)/3(1 \backslash 3) x ^ {\wedge} (- 2 / 3) y ^ {\wedge} (2 / 3) / 6 = (2 \backslash 3) x ^ {\wedge} (1 / 3) y ^ {\wedge} (- 1 / 3) / 36x+3y=7206 x + 3 y = 7 2 0


II. Assume u function of a consumers is u=x(2)+y(2)\mathrm{u} = \mathrm{x}^{\wedge}(2) + \mathrm{y}^{\wedge}(2) then, Required.

1. MRSy,x


Mux=2x\mathrm {M u x} = 2 \mathrm {x}Muy=2y\mathrm {M u y} = 2 \mathrm {y}MRSy,x=Muy/Mux=2y/2x=y/x\mathrm {M R S y , x} = \mathrm {M u y / M u x} = 2 \mathrm {y / 2 x} = \mathrm {y / x}


III. a consumers u function give as u=x(4)y(2)\mathrm{u} = \mathrm{x}^{\wedge}(4) \mathrm{y}^{\wedge}(-2)

1. Mux = 4x^(3) y^(-2)

2. MUy = -2 x^(4) y^(-3)

3. MRSx,y = 4x^(3) y^(-2) : (-2) x^(4) y^(-3) = -2y/x

Answer provided by https://www.AssignmentExpert.com

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS