Question

The value of the rice grows according to the function, 𝑉=𝑘𝑒2√𝑡, how long should the dealer store the rice.

Answer 1

Given

$V=Ke^{2\sqrt t}$

Rice can be stored in a way it maximized the present value of rice, if the interest rate is r.

 Then, the present value is given as:

$P=V.e^{-rt}$

$P=Ke^{2\sqrt {t}} \times e^{-rt}$

Apply log on both sides as follows: 

$lnP=lnk+2\sqrt t-rt$

$\frac{\delta in P}{\delta t}=\frac{2}{2\sqrt t}-r=0$

$\frac{1}{\sqrt t}=r$

$\frac{1}{r}=\sqrt t$

Optimal time $t^*=\frac{1}{r^2}$

In the second order derivative

$\frac{\delta ^2 v}{\delta ^2 t}=v\frac{\delta}{\delta t}(t^{\frac{-1}{2}}-r)\frac{\delta v}{\delta t}$

Here, the product rule is used;

Then, the second term vanishes when: $\frac{\delta v}{\delta t}=0$

Thus,

$\frac{\delta v}{\delta t}=\frac{-v}{2}\sqrt{t^3}<0$

This satisfies the second order condition for maximum.

Thus the dealer should store the rice up to $t=\frac{1}{r^2}$