1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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: $$ \begin{align*} &\frac{-\dfrac{X}{(L+\beta)^2 K} - \dfrac{1}{L}}{-\dfrac{X}{(L+\beta)^2(K+\alpha)}-\dfrac{1}{L + \beta}},\\ &=\frac{\dfrac{-XL-(L+\beta)^2 K}{(L + \beta)^2 KL}}{-\dfrac{X}{X(L+\beta)}-\dfrac{1}{L+\beta}},\\ &= \frac{\dfrac{-(K + \alpha)(L + \beta)L - (L+\beta)^2 K}{(L + \beta)^2KL}}{-\dfrac{2}{(L+\beta)}},\\ &= \dfrac{(K+\alpha)(L + \beta)L+(L+\beta)^2K}{2(L+\beta)KL},\\ &= \dfrac{(K+\alpha)L + (L + \beta)K}{2KL},\\ &= \dfrac{\alpha L+ \beta K + 2 KL}{2KL},\\ &= \dfrac{\alpha L + \beta K}{2KL} + 1 \end{align*} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.