Help with a proof for an quite intuitive Utility optimization problem

Question

Assume $U(x,y,a,c )= - c x + B(x,y,a)$, with $\frac{\partial B(x,y,a)}{\partial c }=0$, and with $a$ and $c\geq 0$ being parameters, and with $x$ and $y$ being variables. Further, $B(x,y,a)$ is continuous and twice continuously differentiable in all its inputs. Further, the $(x,y)$ maximizing $U(x,y,a,c )$ is unique everywhere and it is continuous in $a$ and $c$.

Suppose that $(x_1,y_1)$ maximizes $U(x,y,a_1,c )$, and $(x_2,y_2)$ maximizes $U(x,y,a_2,c )$, and that $0\leq x_1<x_2$.

Let $U^*(a):=\underset{x,y}{\max} U(x,y,a,c)$.

To show: $\frac{\partial U^*(a_1)}{\partial c } > \frac{\partial U^*(a_2)}{\partial c } $

Answer 1

It seems that you can just define two objective problems with the features you stated above and just take the derivatives with respect to $c$ to illustrate this.

Under your specification we have two value functions:

$$U(x_1,y_1,a_1,c)=-cx_1+B(x_1,y_1,a_1)$$ $$U(x_2,y_2,a_2,c)=-cx_2+B(x_2,y_2,a_2)$$

Taking the derivative of the above two equations with respect to $c$ we get:

$$\frac{\partial U(x_1,y_1,a_1,c)}{\partial c}=-x_1$$ $$\frac{\partial U(x_2,y_2,a_2,c)}{\partial c}=-x_2$$

Based on the assumption that $0\leq x_1<x_2$ it follows that $\frac{\partial U(x_1,y_1,a_1,c)}{\partial c}>\frac{\partial U(x_2,y_2,a_2,c)}{\partial c}$.