Science:Math Exam Resources/Courses/MATH104/December 2010/Question 06

MATH104 December 2010

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Question 06

At a certain time, the stock of a company listed on the Vancouver Stock Exchange is selling for $0.25 a share. The controlling shareholders spread a (quickly dispelled) rumour that the company has discovered a cure for death. This causes a surge of demand for the stock. As a result, when $t\geq {0}$ the price of a share is P(t) dollars, where $\displaystyle {}P(t)=0.25+30t{\textrm {e}}^{-t/5}.$ Here t is measured in days and t = 0 at the instant the rumour starts. At what time t is the price of the stock increasing most rapidly? Please justify that the answer you calculate actually gives the greatest rate of increase.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
What is the objective function (the thing you are trying to maximize)? (Hint: It is not the price of the stock.)

Hint 2
If we find a critical point for the objective function, how can we use derivative information to conclude if it's a maximum or a minimum? If it is not a maximum, where else do we look to maximize the objective function?

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Step 1: Find objective function We have that the stock price is $\displaystyle {}P(t)=0.25+30t{\textrm {e}}^{-t/5}$ and are asked to find when the stock is increasing most rapidly. Notice we are not asked to find when the price of a stock is a maximum. This makes sense since if you're looking to buy the stock, you want it when its value is increasing the fastest. If you buy the stock at its maximum price then your stock will only lose value as time moves forward. Therefore the objective function is to maximize the rate of change, R(t), of the stock price or $\displaystyle {}P'(t)$. We get, $\displaystyle {}R(t)=P'(t)=30{\textrm {e}}^{-t/5}-6t{\textrm {e}}^{-t/5}=(30-6t){\textrm {e}}^{-t/5}.$ Step 2: Find critical points of the objective function R(t) We start by finding critical points of R(t). First take the derivative of R(t) and set it to zero. $R'(t)=-6{\textrm {e}}^{-t/5}-{\frac {1}{5}}(30-6t){\textrm {e}}^{-t/5}=\left(-6-{\frac {1}{5}}(30-6t)\right){\textrm {e}}^{-t/5}=\left({\frac {6}{5}}t-12\right){\textrm {e}}^{-t/5}=0.$ Since the exponential can never be zero we must have that, ${\begin{aligned}\left({\frac {6}{5}}t-12\right)&=0\\t&=10.\end{aligned}}$ Thus t=10 is the only critical point of R(t). Step 3: Is the critical point a maximum or a minimum? We have still not answered the question, we must still maximize R. We have that t is in $[0,\infty )$ and so we do not have a full closed interval to use the extreme value theorem. However, since we only have one critical point we know that if R(10) is a local maximum then it is also an absolute maximum and if R(10) is a local minimum then it is also an absolute minimum. We will use the first derivative test. Notice, if we pick something slightly less than 10, then R'(t)<0 so R(t) is decreasing. Similarly if we pick something slightly bigger than 10 then R'(t)>0 so R(t) is increasing. Therefore, by the first derivative test we conclude that t=10 is a local minimum. Since it is the only critical point, it is also an absolute minimum. Step 4: Look at the boundaries of the interval of t to find the absolute maximum How do we then find the absolute maximum? Even though we do not have a closed interval for R(t), we can see what happens for large time, i.e. we can compute $\lim _{t\to \infty }R(t).$ Doing so involves taking an infinite limit of $t{\textrm {e}}^{-t/5}$. The exponential decays much faster than the linear term grows so this term goes to zero. Therefore we have that, ${\begin{aligned}&\lim _{t\to \infty }R(t)\\&=\lim _{t\to \infty }30{\textrm {e}}^{-t/5}-6\lim _{t\to \infty }6t{\textrm {e}}^{-t/5}\\&=0.\end{aligned}}$ Therefore we see that for large time, R(t) goes to zero. We can plug in the left endpoint, $\displaystyle {}R(0)=(30-6(0)){\textrm {e}}^{-0/5}=30.$ Step 5: Conclusion Therefore, we see that R starts at 30, decreases to the critical point at t=10 (note that R(10) < 0) and then increases towards 0 for t large. Therefore, R is the biggest when t=0, or right when the rumour starts spreading. Therefore, the price of the stock is increasing most rapidly when t=0.

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