Question #189254

Consider the production function Q = 2(KL)0.5

a)     What is the marginal product of labour and capital

b)     What is the marginal rate of technical substitution of labor for capital


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


 

1
Expert's answer
2021-05-06T15:53:20-0400

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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