Science:Math Exam Resources/Courses/MATH104/December 2010/Question 06/Solution 1

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.