# Tag Info

You have that in equilibrium each factor is paid its marginal product, so $$\tag1\frac{w_i}{p_i}=B_if'(l_i)$$ and $$\tag2\frac{r_i}{p_i}=f(l_i)-K_if'(l_i)\frac{B_iL_i}{K_i^2}=f(l_i)-l_if'(l_i)$$ so deviding (2) by (1) we have: $$\tag3\frac{r_i}{w_i}=\frac{f(l_i)-l_if'(l_i)}{B_if'(l_i)}.$$ Take the derivative of $r_i/w_i$ with respect to $l_i$: \tag4\...