<p>It seems that you can just define two objective problems with the features you stated above and just take the derivatives with respect to <span class="math-container">$c$</span> to illustrate this.</p>
<p>Under your specification we have two value functions:</p>
<p><span class="math-container">$$U(x_1,y_1,a_1,c)=-cx_1+B(x_1,y_1,a_1)$$</span>
<span class="math-container">$$U(x_2,y_2,a_2,c)=-cx_2+B(x_2,y_2,a_2)$$</span></p>
<p>Taking the derivative of the above two equations with respect to <span class="math-container">$c$</span> we get:</p>
<p><span class="math-container">$$\frac{\partial U(x_1,y_1,a_1,c)}{\partial c}=-x_1$$</span>
<span class="math-container">$$\frac{\partial U(x_2,y_2,a_2,c)}{\partial c}=-x_2$$</span></p>
<p>Based on the assumption that <span class="math-container">$0\leq x_1&lt;x_2$</span> it follows that <span class="math-container">$\frac{\partial U(x_1,y_1,a_1,c)}{\partial c}&gt;\frac{\partial U(x_2,y_2,a_2,c)}{\partial c}$</span>.</p>

CC BY-SA 4.0
2023-10-13T07:47:17.627Z
+1 but maybe you can add that you use the envelope theorem to obtain the expression of the partial derivatives.
tdm
CC BY-SA 4.0
2023-10-13T09:32:17.190Z
@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
CC BY-SA 4.0
2023-10-13T09:45:42.260Z
Combing it with @tdm: The envelope theorem says that we can "ignore" $x^*$'s and $y^*$ change in $c$.
Paul
CC BY-SA 4.0
2023-10-15T00:10:24.500Z
@Paul, Correct there's some envelope theorem applied here implicitly.
EconJohn
CC BY-SA 4.0
2023-10-12T19:25:02.813Z
2023-10-12T19:25:02.813Z
EconJohn
<p>Assume <span class="math-container">$U(x,y,a,c )= - c x + B(x,y,a)$</span>, with <span class="math-container">$\frac{\partial B(x,y,a)}{\partial c }=0$</span>, and with <span class="math-container">$a$</span> and <span class="math-container">$c\geq 0$</span> being parameters, and with <span class="math-container">$x$</span> and <span class="math-container">$y$</span> being variables. Further, <span class="math-container">$B(x,y,a)$</span> is continuous and twice continuously differentiable in all its inputs. Further, the <span class="math-container">$(x,y)$</span> maximizing <span class="math-container">$U(x,y,a,c )$</span> is unique everywhere and it is continuous in <span class="math-container">$a$</span> and <span class="math-container">$c$</span>.</p>
<p>Suppose that <span class="math-container">$(x_1,y_1)$</span> maximizes <span class="math-container">$U(x,y,a_1,c )$</span>, and <span class="math-container">$(x_2,y_2)$</span> maximizes <span class="math-container">$U(x,y,a_2,c )$</span>, and that <span class="math-container">$0\leq x_1&lt;x_2$</span>.</p>
<p>Let <span class="math-container">$U^*(a):=\underset{x,y}{\max} U(x,y,a,c)$</span>.</p>
<p>To show: <span class="math-container">$\frac{\partial U^*(a_1)}{\partial c } &gt; \frac{\partial U^*(a_2)}{\partial c }  $</span></p>

<microeconomics><optimization>
Help with a proof for an quite intuitive Utility optimization problem
Paul
# Help with a proof for an quite intuitive Utility optimization problem

(Asked by: username_0 on 2023-10-10)

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 } $

**Tags:** microeconomics, optimization

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## 1 answers

### Answer by username_1 on 2023-10-12. Score: 3 | Accepted answer

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 $\frac{\partial U(x\_1,y\_1,a\_1,c)}{\partial c}>\frac{\partial U(x\_2,y\_2,a\_2,c)}{\partial c}$.

> **Comments:**
> **username_2**: +1 but maybe you can add that you use the envelope theorem to obtain the expression of the partial derivatives.  
> **username_0**: @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?  
> **username_0**: Combing it with @tdm: The envelope theorem says that we can "ignore" $x^*$'s and $y^*$ change in $c$.  
> **username_1**: @Paul, Correct there's some envelope theorem applied here implicitly.  

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