<p>Note that, from your SOC:
<span class="math-container">\begin{align}
P''(t) &amp;= V''(t)e^{-rt} -  rV'(t)e^{-rt}-  rV'(t)e^{-rt} + r^2V(t)e^{-rt} + rme^{-rt}\\
&amp;=e^{-rt}\Bigl[\underbrace{\color{red}{V''(t)-rV'(t)}}_{=D}-r\underbrace{\bigl[V'(t)-rV(t)-m\bigr]}_{=0\text{ from FOC}}\Bigr]
\end{align}</span></p>

CC BY-SA 4.0
2021-02-16T09:36:22.727Z
Thanks, it makes sense I guess. But I don't understand why then we don't include $e^{-rt}$ in D, I know it has to be positive but is it proper to just say that SOC are given by D just because  $e^{-rt}>0$?
WilliamT
CC BY-SA 4.0
2021-02-16T15:28:58.043Z
@WilliamT: You're right, and I think the author of the answer key was just being sloppy.
Herr K.
CC BY-SA 4.0
2021-02-16T02:20:10.530Z
2021-02-16T02:20:10.530Z
Herr K.
<p>I am trying to find first and second order conditions for the following problem:</p>
<p><span class="math-container">$$P(t) = V(t)e^{-rt}- \int^t_0 me^{-r\tau}d \tau $$</span></p>
<p>I managed to find the first order condition:</p>
<p><span class="math-container">$$P'(t) = V'(t)e^{-rt} -  rV(t)e^{-rt} - me^{-rt} =0 \\
V'(t) = rV(t)+m $$</span></p>
<p>According to our textbook this is correct, but I have problem finding SOC.  We learned before that SOC are found by taking second order derivative and examining if it is &gt; or &lt; than 0.</p>
<p>When I try to find <span class="math-container">$P''(t)$</span> I get:</p>
<p><span class="math-container">$$P''(t) = V''(t)e^{-rt} -  rV'(t)e^{-rt}-  rV'(t)e^{-rt} + r^2V(t)e^{-rt} + rme^{-rt}$$</span></p>
<p>But the textbook says that the second order condition should be:</p>
<p><span class="math-container">$$D=V''(t) -rV'(t)$$</span></p>
<p>this looks like the derivative of <span class="math-container">$V'(t) = rV(t)+m \implies V'(t) - rV(t) - m = 0 $</span></p>
<p>but I don't understand why that is correct. Previously when deriving SOC we always just took the 2nd derivative of the function so how come that does not work here?</p>

CC BY-SA 4.0
2021-02-15T19:38:30.040Z
Your formula for $P''(t)$ looks correct.  To find whether it is > or < 0 at the point where $P'(t)=0$, you can substitute $V'(t)=rV(t)+m$ and also note that $e^{-rt}>0$.
Adam Bailey
CC BY-SA 4.0
2021-02-15T22:23:49.680Z
hi, thanks. But why then the textbook says the SOC is given by: $V′′(t)−rV′(t)>0$? I guess I was probably not clear in my question, I know how to check $P''(t)$, but the credited answer is different and I can't believe this is typo since it is too different
WilliamT
CC BY-SA 4.0
2021-02-16T08:51:21.277Z
Can I ask your textbook out of curiosity?
EB3112
CC BY-SA 4.0
2021-02-16T09:33:46.297Z
@EB3112 it is Essential Mathematics for Economic Analysis by Hammond and other authors
WilliamT
<microeconomics><mathematical-economics>
Finding optimum market value of vintage car
WilliamT
# Finding optimum market value of vintage car

(Asked by: username_0 on 2021-02-15)

I am trying to find first and second order conditions for the following problem:

$$P(t) = V(t)e^{-rt}- \int^t\_0 me^{-r\tau}d \tau $$

I managed to find the first order condition:

$$P'(t) = V'(t)e^{-rt} - rV(t)e^{-rt} - me^{-rt} =0 \\
V'(t) = rV(t)+m $$

According to our textbook this is correct, but I have problem finding SOC. We learned before that SOC are found by taking second order derivative and examining if it is > or < than 0.

When I try to find $P''(t)$ I get:

$$P''(t) = V''(t)e^{-rt} - rV'(t)e^{-rt}- rV'(t)e^{-rt} + r^2V(t)e^{-rt} + rme^{-rt}$$

But the textbook says that the second order condition should be:

$$D=V''(t) -rV'(t)$$

this looks like the derivative of $V'(t) = rV(t)+m \implies V'(t) - rV(t) - m = 0 $

but I don't understand why that is correct. Previously when deriving SOC we always just took the 2nd derivative of the function so how come that does not work here?

**Tags:** microeconomics, mathematical-economics
> **Comments:**
> **username_1**: Your formula for $P''(t)$ looks correct.  To find whether it is > or < 0 at the point where $P'(t)=0$, you can substitute $V'(t)=rV(t)+m$ and also note that $e^{-rt}>0$.  
> **username_0**: hi, thanks. But why then the textbook says the SOC is given by: $V′′(t)−rV′(t)>0$? I guess I was probably not clear in my question, I know how to check $P''(t)$, but the credited answer is different and I can't believe this is typo since it is too different  
> **username_2**: Can I ask your textbook out of curiosity?  
> **username_0**: @EB3112 it is Essential Mathematics for Economic Analysis by Hammond and other authors  

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## 1 answers

### Answer by username_3 on 2021-02-16. Score: 4 | Accepted answer

Note that, from your SOC:
\begin{align}
P''(t) &= V''(t)e^{-rt} - rV'(t)e^{-rt}- rV'(t)e^{-rt} + r^2V(t)e^{-rt} + rme^{-rt}\\
&=e^{-rt}\Bigl[\underbrace{\color{red}{V''(t)-rV'(t)}}\_{=D}-r\underbrace{\bigl[V'(t)-rV(t)-m\bigr]}\_{=0\text{ from FOC}}\Bigr]
\end{align}

> **Comments:**
> **username_0**: Thanks, it makes sense I guess. But I don't understand why then we don't include $e^{-rt}$ in D, I know it has to be positive but is it proper to just say that SOC are given by D just because  $e^{-rt}>0$?  
> **username_3**: @WilliamT: You're right, and I think the author of the answer key was just being sloppy.  

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