<p>Note that your problem said to assume that <span class="math-container">$d &lt; 0$</span>. When you divide both sides of an inequality by a negative number, you flip the signs. For example:
<span class="math-container">$$ -2x &lt; 4 \Rightarrow x &gt; -2 $$</span></p>
<p>Once you handle that, you can see that both methods imply the same relationship.</p>
<p>Levels: <a href="https://i.sstatic.net/2CAuo.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/2CAuo.png" alt="U(W)" /></a></p>
<p>First derivative: <a href="https://i.sstatic.net/u5RMA.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/u5RMA.png" alt="U'(W)" /></a></p>

CC BY-SA 4.0
2022-08-02T15:05:57.717Z
2022-08-02T15:05:57.717Z
BKay
<p><strong>Question</strong></p>
<p>A risk-averse, non-satiated investor has decided to use the utility function <span class="math-container">$$U(w) = w + dw^2,$$</span> where <span class="math-container">$$d \leq 0$$</span> is a constant, to describe his preferences.</p>
<p>Derive an upper bound for <span class="math-container">$w$</span>, in terms of <span class="math-container">$d$</span>, for which <span class="math-container">$U(w)$</span> is valid.</p>
<hr />
<p><strong>My Answer</strong></p>
<p>I know that, for non-satiation, <span class="math-container">$$U'(w) \geq 0.$$</span></p>
<p><em>Approach 1</em></p>
<p><span class="math-container">$$\begin{aligned}
1 + 2dw \geq 0\\
\implies w \geq -\frac 1 {2d}
\end{aligned}$$</span></p>
<p><em>Approach 2</em></p>
<p>Let <span class="math-container">$$e = -d.$$</span>
<span class="math-container">$$\begin{aligned}
1 - 2ew \geq 0\\
\implies w \leq \frac 1 {2e}\\
\implies w \leq -\frac 1 {2d}
\end{aligned}$$</span></p>
<hr />
<p>Since the question is asking for an upper bound, I deduce that my second approach is the correct one. However, I would like to know why the first approach is incorrect.</p>
<p>Any intuitive explanations will be greatly appreciated :)</p>

CC BY-SA 4.0
2022-08-02T14:48:53.877Z
You need to track the sign of $d$ whether it's implicit (as in the first approach) or explicit (as in the second) when dividing. So even in the first, the inequality switches direction and you get the same result both ways.
BrsG
CC BY-SA 4.0
2022-08-02T14:50:44.930Z
Voting to close: The solution is unrelated to economics, but simply an application of rather basic math. Not worthwhile documenting here.
BrsG
<utility><preferences><wealth>
On quadratic utility functions
Ethan Mark
# On quadratic utility functions

(Asked by: username_0 on 2022-08-02)

**Question**

A risk-averse, non-satiated investor has decided to use the utility function $$U(w) = w + dw^2,$$ where $$d \leq 0$$ is a constant, to describe his preferences.

Derive an upper bound for $w$, in terms of $d$, for which $U(w)$ is valid.

---

**My Answer**

I know that, for non-satiation, $$U'(w) \geq 0.$$

*Approach 1*

$$\begin{aligned}
1 + 2dw \geq 0\\
\implies w \geq -\frac 1 {2d}
\end{aligned}$$

*Approach 2*

Let $$e = -d.$$
$$\begin{aligned}
1 - 2ew \geq 0\\
\implies w \leq \frac 1 {2e}\\
\implies w \leq -\frac 1 {2d}
\end{aligned}$$

---

Since the question is asking for an upper bound, I deduce that my second approach is the correct one. However, I would like to know why the first approach is incorrect.

Any intuitive explanations will be greatly appreciated :)

**Tags:** utility, preferences, wealth
> **Comments:**
> **username_1**: You need to track the sign of $d$ whether it's implicit (as in the first approach) or explicit (as in the second) when dividing. So even in the first, the inequality switches direction and you get the same result both ways.  
> **username_1**: Voting to close: The solution is unrelated to economics, but simply an application of rather basic math. Not worthwhile documenting here.  

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## 1 answers

### Answer by username_2 on 2022-08-02. Score: 2 | Accepted answer

Note that your problem said to assume that $d < 0$. When you divide both sides of an inequality by a negative number, you flip the signs. For example:
$$ -2x < 4 \Rightarrow x > -2 $$

Once you handle that, you can see that both methods imply the same relationship.

Levels: [![U(W)](https://i.sstatic.net/2CAuo.png)](https://i.sstatic.net/2CAuo.png)

First derivative: [![U'(W)](https://i.sstatic.net/u5RMA.png)](https://i.sstatic.net/u5RMA.png)


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