Question

the abercrombie and fitch stock had a price given as $p = 0.2t^2 - 5.6t + 50.2$, where $t$ is the time in months from 1999 to 2001. ( $t = 1$ is january 1999). find the two months in which the price of the stock was $\\$30$.

a. april 1999 and january 2001
b. april 1999 and february 2000
c. april 1999 and december 2000
d. april 1999 and february 2001

Explanation:

Step1: Set $P=30$, rearrange equation

$0.2t^2 - 5.6t + 50.2 = 30$
$0.2t^2 - 5.6t + 20.2 = 0$
Multiply all terms by 10 to eliminate decimals:
$2t^2 - 56t + 202 = 0$
Divide by 2 to simplify:
$t^2 - 28t + 101 = 0$

Step2: Apply quadratic formula

Quadratic formula: $t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, where $a=1$, $b=-28$, $c=101$
First calculate discriminant:
$\Delta = (-28)^2 - 4(1)(101) = 784 - 404 = 380$
$\sqrt{\Delta} = \sqrt{380} \approx 19.49$
Solve for $t$:
$t_1 = \frac{28 + 19.49}{2} \approx \frac{47.49}{2} \approx 23.75$
$t_2 = \frac{28 - 19.49}{2} \approx \frac{8.51}{2} \approx 4.26$

Step3: Map t to months

$t\approx4$: $t=1$ is Jan 1999, so $t=4$ is April 1999
$t\approx24$: $t=12$ is Dec 1999, $t=24$ is Dec 2000? No, $t=25$ is Jan 2001, so $t\approx23.75$ is late Dec 2000, which is December 2000. Wait, correction: $t=1$ Jan 1999, $t=13$ Jan 2000, $t=25$ Jan 2001. So $t=23$ is Nov 2000, $t=24$ is Dec 2000. $t\approx23.75$ is Dec 2000.

Answer:

C. April 1999 and December 2000