# Help with a proof for an quite intuitive Utility optimization problem

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.

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

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

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

• +1 but maybe you can add that you use the envelope theorem to obtain the expression of the partial derivatives.
– tdm
Oct 13 at 7:47
• @EconJohn: Thanks for your answer. The problem is that $x_1$ and $x_2$ are also functions of $c$. Why can we ignore them when taking the derivative?
– Paul
Oct 13 at 9:32
• Combing it with @tdm: The envelope theorem says that we can "ignore" $x^*$'s and $y^*$ change in $c$.
– Paul
Oct 13 at 9:45
• @Paul, Correct there's some envelope theorem applied here implicitly.
– EconJohn
Oct 15 at 0:10
• Thanks to both of you!
– Paul
Oct 16 at 11:26