For this function, the marginal rate of technical substitution is given by (K+alpha)/(L+beta). Generally we solve for K/L in terms of MRTS of two factors. Then differentiate to solve for elasticity of substitution. In this case, K/L is becoming a function of both MRTS and L also. How to approach the problem in this case.


1 Answer 1


Touch wood that I did not make any mistakes.

Consider the production function $X = (K + \alpha)(L + \beta)$.

The elasticity of substitution is given by: $$ \frac{\partial \ln(K/L)}{\partial\ln(MP_L/MP_K)} = \frac{\partial(\ln(K) - \ln(L)}{\partial(\ln(K + \alpha)- \ln(L + \beta))} $$ Let's take the derivative of both numerator and denominator with respect to $L$, writing $K$ as a function of $L$ itself: $$ \dfrac{\dfrac{K_L}{K}- \dfrac{1}{L}}{\dfrac{K_L}{K + \alpha} - \dfrac{1}{L + \beta}},\\ $$ Now, rewriting the production function, we have that: $K = \dfrac{X}{L + \beta} - \alpha$, so keeping $X$ fixed we have: $$ K_L = -\frac{X}{(L+\beta)^2} $$ Substitution gives: $$ \begin{align*} &\frac{-\dfrac{X}{(L+\beta)^2 K} - \dfrac{1}{L}}{-\dfrac{X}{(L+\beta)^2(K+\alpha)}-\dfrac{1}{L + \beta}},\\ &=\frac{\dfrac{-XL-(L+\beta)^2 K}{(L + \beta)^2 KL}}{-\dfrac{X}{X(L+\beta)}-\dfrac{1}{L+\beta}},\\ &= \frac{\dfrac{-(K + \alpha)(L + \beta)L - (L+\beta)^2 K}{(L + \beta)^2KL}}{-\dfrac{2}{(L+\beta)}},\\ &= \dfrac{(K+\alpha)(L + \beta)L+(L+\beta)^2K}{2(L+\beta)KL},\\ &= \dfrac{(K+\alpha)L + (L + \beta)K}{2KL},\\ &= \dfrac{\alpha L+ \beta K + 2 KL}{2KL},\\ &= \dfrac{\alpha L + \beta K}{2KL} + 1 \end{align*} $$


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