Under what conditions would a quasilinear utility function in a function form exhibit diminishing marginal rate of substitution?

Question

Let the utility function be: $U(x_1,x_2) = x_1 + x_2^\alpha$. Diminishing MRS requires $\frac{dMRS}{dx_1} <0$, however, taking this derivative results in 0, as $MRS = \frac{1}{\alpha x_2^{\alpha -1}}$.

1. How do we interpret this result?

2. Does diminishing MRS hold when the second order condition of $x_1$ is 0?

3. $\frac{dMRS}{dx_2} = \frac{1-\alpha}{\alpha x_2^\alpha} > 0$, when $\alpha < 1$, is this enough to confirm diminishing MRS?

4. Can $\alpha$ be negative? Does that make sense in a utility function?

Answer 1

Normally, decreasing MRS means that the MRS decreases as one moves along the indifference curve in the direction of $x$. As you compute it, you increase $x$ but you do not move along the indifference curve because you do not keep the utility level constant.

Consider the utility level $U$. Then the equation for the indifference curve with level $U$ is: $$ x_1 + x_2^\alpha = U \to x_2 = (U - x_1)^{1/\alpha} $$ (note that we only evaluate the right hand side over $x_1 \le U$ for $x_2$ not to become negative).

The MRS is then: $$ MRS = \left|\frac{dx_2}{dx_1}\right| = \left|-\frac{1}{\alpha}(U - x_1)^{\frac{1 - \alpha}{\alpha}}\right| = \frac{1}{\alpha}(U - x_1)^{\frac{1 - \alpha}{\alpha}}. $$ Now, if you evaluate this at $U = x_1 + x_2^\alpha$, you indeed obtain that: $$ \left|\frac{dx_2}{dx_1}\right|_{U = x_1 + x_2^\alpha} = \frac{1}{\alpha x_2^{\alpha - 1}}. $$ However, the change in the MRS due to a change in $x_1$ along the indifference curve with utility value $U$ is: $$ \frac{1}{dx_1}\left|\frac{d x_2}{d x_1}\right| = -\frac{(1 - \alpha)}{\alpha^2}(U - x_1)^{\frac{1 - 2 \alpha}{\alpha}}. $$ which is negative for $\alpha < 1$ (note that $x_1 \le U$).