AP Calculus AB 5.5 Absolute Extrema

[Maximum number: 3]

An invasive species of plant appears in a fruit grove at time t=0 and begins to spread. The function C defined by $C(t)=7.6 \arctan (0.2 t)$ models the number of acres in the fruit grove affected by the species t weeks after the species appears. It can be shown that $C^{\prime}(t)=\frac{38}{25+t^{2}}$ .

(Note: Your calculator should be in radian mode.)

At time t=4 weeks after the invasive species appears in the fruit grove, measures are taken to counter the spread of the species. The function A, defined by $A(t)=C(t)-\int_{4}^{t} 0.1 \cdot \ln (x) d x$ , models the number of acres affected by the species over the time interval $4 \leq t \leq 36$ . At what time t, for $4 \leq t \leq 36, \operatorname{does} A$ attain its maximum value? Justify your answer.

D At time t=4 weeks after the invasive species appears in the fruit grove, measures are taken to counter

the spread of the species. The function A, defined by $A(t)=C(t)-\int_{4}^{t} 0.1 \cdot \ln (x) d x$ , models the number

of acres affected by the species over the time interval $4 \leq t \leq 36$ . At what time t, for $4 \leq t \leq 36$ , does

A attain its maximum value? Justify your answer.

$A^{\prime}(t)=C^{\prime}(t)-0.1 \cdot \ln t$	Considers $A^{\prime}(t)=0 \quad$ Point 7 (P7)
For $4 \leq t \leq 36$ , the maximum value of A(t) occurs when $A^{\prime}(t)=0$ or at an endpoint. $A^{\prime}(t)=C^{\prime}(t)-0.1 \cdot \ln t=0 \Rightarrow C^{\prime}(t)=0.1 \cdot \ln t$
$\Rightarrow t=11.441700$ t	A(t)<br>4	5.128031<br>11.441700	7.316978<br>36	1.743056	Justification Point 8 (P8)
Therefore, the number of acres affected by the species is a maximum at time t=11.442 (or 11.441) weeks.	Answer with supporting work

Scoring Notes for Part D

- P7 is earned for considering $A^{\prime}(t)=0, C^{\prime}(t)-0.1 \cdot \ln t=0$ , or $C^{\prime}(t)=0.1 \cdot \ln t$ . P7 is not earned

by just presenting t=11.441700.

A response that discusses the sign of $A^{\prime}(t)$ changing or uses the phrase "critical points of A " also

earns P7.

- To earn P8 using a candidates test, a response must make a global argument by correctly evaluating

A(t) at t=4, t=11.441700, and t=36. The evaluations must be correct to the first digit after

the decimal, rounded or truncated.

- Alternate justifications:

○ $A^{\prime}(t)>0$ for 4<t<11.442, and $A^{\prime}(t)<0$ for 11.442<t<36. Therefore, t=11.442 is the

location of the absolute maximum for A on the interval $4 \leq t \leq 36$ .

○ Because $A^{\prime}(t)$ changes sign from positive to negative at t=11.442 (this might be presented as

" $A^{\prime}(t)>0$ for t<11.442, and $A^{\prime}(t)<0$ for t>11.442 "), it is the location of a relative

maximum for A. And because t=11.442 is the only critical point of A in the interval

$4 \leq t \leq 36$ , it is the location of the absolute maximum for A on the interval.

- A response that presents a local argument (such as a First Derivative Test or a Second Derivative

Test) or an incorrect global argument does not earn P8 but is eligible for P9 with the correct answer.

A reported answer should be accurate to three places after the decimal point, rounded or truncated.

An inappropriately rounded answer does not earn the point, unless an earlier point was not earned

due to inappropriate rounding.

Part A (AB): Graphing calculator required

AP Calculus AB Fun 4 A Justify Conclusions About the Behavior of a Function Based on the Behavior of Its Derivatives Topic 5 5 Questions question 1