Question #188396

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


1
Expert's answer
2021-05-04T11:15:54-0400

Given

V=Ke2tV=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.ertP=V.e^{-rt}

P=Ke2t×ertP=Ke^{2\sqrt {t}} \times e^{-rt}

Apply log on both sides as follows: 

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


δinPδt=22tr=0\frac{\delta in P}{\delta t}=\frac{2}{2\sqrt t}-r=0


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


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


Optimal time t=1r2t^*=\frac{1}{r^2}


In the second order derivative

δ2vδ2t=vδδt(t12r)δvδt\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: δvδt=0\frac{\delta v}{\delta t}=0

Thus,

δvδt=v2t3<0\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=1r2t=\frac{1}{r^2}


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