# Regression analysis of the price of stock gives the price in month t as P(t) = t^3/10 - 3t + 10. An investor has \$1200 to be invested and can stay invested for 3 years. What is the maximum gain that can be made.

Tushar Chandra | Certified Educator

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It has been determined from statistical analysis of the price of a stock that its value is a function of the month (t) and is given by P(t) = (1/10)*t^3 - 3t + 10.

To maximize profit the investor should buy the stock at the lowest price it is available at and sell it when the price is at the highest level. (Here, profit that can be made by short selling the stock is not being considered)

The function P(t) = (1/10)*t^3 - 3t + 10 can be used to determine the lowest and highest price. First, determine the derivative of P(t). This is P'(t) = (3/10)*t^2 - 3. Solving P'(t) = 0 gives t = +sqrt(10) and t = -sqrt(10). Ignore the negative value of t. At t = sqrt 10, the value of P''(t) is positive indicating that at this point the price is at the lowest level. The lowest price at which the stock can be bought is P(sqrt 10) = (sqrt 10)^3/10 - 3*sqrt 10 + 10 = 3.675

Once the lowest price of \$3.675 is reached, the price of the stock continues to rise. As the investor can stay invested for 3 years or 36 months, the highest price at which the stock can be sold is (36)^3/10 - 3*36 + 10 = \$4567.6

At \$3.675, the investor can buy 1200/3.675 = 326 stocks of the company. When they are sold at 4567.6, the amount received is \$1489037.6.

The profit that can be made is equal to difference between the final selling price and the initial buying price. This is \$1489037.6 - \$1200 = \$1487837.6

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