Need help understanding with the following? Find the elasticity of scale and the elasticity of substitution for the CES production function: f(x_1, x_2 )=(x_1^(1/3)+x_2^(1/3))^3.

Solution:

We first calculate the marginal products:

f_(x_1 )=3(x_1^(1/3)+x_2^(1/3) )^2 (1/3 x_1^(-2/3) )= (x_1^(1/3)+x_2^(1/3) )^2 x_1^(-2/3)

f_(x_2=) 3(x_1^(1/3)+x_2^(1/3) )^2 (1/3 x_2^(-2/3) )= (x_1^(1/3)+x_2^(1/3) )^2 x_2^(-2/3)

Elasticity of scale = (x_1 f_(x_1 ))/(f(x_1, x_2))+(x_2 f_(x_2 ))/(f(x_1, x_2))=(x_1 (x_1^(1/3)+x_2^(1/3) )^2 x_1^(-2/3))/(x_1^(1/3)+x_2^(1/3))^3 + (x_2 (x_1^(1/3)+x_2^(1/3) )^2 x_2^(-2/3))/(x_1^(1/3)+x_2^(1/3))^3 =(x_1^(1/3))/((x_1^(1/3)+x_2^(1/3) ) )+(x_2^(1/3))/((x_1^(1/3)+x_2^(1/3)))=1.

To get elasticity of substitution, we first need TRS and denote r=x_2/x_1 .

TRS=f_(x_1 )/f_(x_2 ) =((x_1^(1/3)+x_2^(1/3) )^2 x_1^(-2/3))/((x_1^(1/3)+x_2^(1/3) )^2 x_2^(-2/3) )=(x_2^(2/3))/(x_1^(2/3) )=r^(2/3)

r^(2/3)=TRS=t (where TRS = t).

r=t^(3/2)ln(r)=3/2 ln(t) and =(dln(r))/(dln(t))=3/2.