Finding optimum market value of vintage car

Question

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?

Answer 1

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}