Question #184191

Consider the production function Q = 2(KL)0.5

What is the marginal product of labour and capital (1 marks)

What is the marginal rate of technical substitution of labor for capital (2 marks)


What is the elasticity of substitution at a point K = 1, L = 1 if we increase K by one unit? (2 marks)


1
Expert's answer
2021-04-23T12:01:55-0400

With the given function of production Q=2(KL)0.5Q=2(KL)^{0.5}

1) To find the products (MPK and also MPL), we use this formula. 

Q=2(KL)0.5Q=2(KL)^{0.5}


MPK=QK=L0.5MPK=\frac{∂Q}{∂K}=L^{0.5}


MPK=L0.5MPK=L^{0.5}


MPL=δQδL=(KL)0.5MPL=\frac{\delta Q}{\delta L}=(KL)^{-0.5}


2) To find the value of substitution (technical) of labor for capital

MRTS=MPLMPKMRTS=\frac {MPL}{MPK}


MPTS=KL0.5L0.5MPTS=\frac{KL^{-0.5}}{L^{0.5}}


MRTS=1KL0.5MRTS=\frac {1}{KL^{0.5}}


3) To find how much the substitution is elastic


ε=(Δ1kΔMRTS)MRTS1kε=(\frac{\Delta \frac{1}{k}}{\Delta MRTS})\frac{MRTS}{\frac{1}{k}}


we know that MRTS=1KL0.5MRTS=\frac {1}{KL^{0.5}}


taking the derivative

MRTS=1KL0.5MRTS=\frac {1}{KL^{0.5}}


MRTS=0.5(KL)1.5MRTS=0.5(KL)^{-1.5}


MRTS=10.5(KL)1.5MRTS=\frac {1}{0.5(KL)^{1.5}}


MRTS=10.5(KL)1.5MRTS=\frac {1}{0.5(KL)^{1.5}}


Δ1kΔMRTS=10.5(KL)1.5\frac{\Delta \frac{1}{k}}{\Delta MRTS}=\frac {1}{0.5(KL)^{1.5}}


ε=10.5KLε=\frac{1}{0.5KL}


ε=10.5(1×1)ε=\frac{1}{0.5(1\times 1)}


ε=2ε=2


the elasticity of substitution would be 2

but, if K is raised by 1, then


ε=10.5KLε=\frac{1}{0.5KL}


ε=10.5(2×1)ε=\frac{1}{0.5(2 \times 1)}


ε=1ε=1


Now the elasticity will fall to 1



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Comments

Jeshmita
12.10.21, 11:52

very helpful.

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