Question

The price of a certain computer stock t days after it is issued for sale is p(t)=100+20t−6t^2 dollars. The price of the stock initially rises, but eventually begins to fall. During what period of time does the stock price rise?
0 < t < ____?
If you owned the stock, after how many days would you sell it? ____ days

Answer 1

Answer:

0 < t < $\frac{5}{3}$

After 1.67 days the stocks would be sold out.

Step-by-step explanation:

The price of a certain computer stock after t days is modeled by

p(t) = 100 + 20t - 6t²

Now we will take the derivative of the given function and equate it to zero to find the critical points,

p'(t) = 20 - 12t = 0

t = $\frac{20}{12}$

t = $\frac{5}{3}$ days

Therefore, there are two intervals in which the given function is defined

(0, $\frac{5}{3}$) and ($\frac{5}{3}$, ∞)

For the interval (0, $\frac{5}{3}$),

p'(1) = 20 - 12(1) = 20

For the interval ($\frac{5}{3}$, ∞),

p'(2) = 20 - 12(2) = -4

Positive value of p'(t) in the interval (0, $\frac{5}{3}$) indicates that the function is increasing.

0 < t < $\frac{5}{3}$

Since at the point t = 1.67 days curve is showing the maximum, so the stocks should be sold after 1.67 days.