Answer to Question #188396 in Economics of Enterprise for faryal

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=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}"


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