Question

PLEASE HELP URGENT : The rate at which customers arrive at a counter to be served is modeled by the function F defined by ​F(t)=10+6cos((t)/(\pi )) 0<=t<=60 where​ F(t) is measured in customers per minute and t is measured in minutes. To the nearest whole​ number, how many customers arrive at the counter over the 60​-minute ​period?

Answer 1

Answer:

606 customers arrive at the counter over the 60 minute period.

Step-by-step explanation:

Given : The rate at which customers arrive at a counter to be served is modeled by the function F defined by ​$F(t)=10+6\cos(\frac{t}{\pi})$ $0\leq t\leq 60$  where​ F(t) is measured in customers per minute and t is measured in minutes.

To find : How many customers arrive at the counter over the 60​-minute ​period?

Solution :

We are going to integrate the function in the interval $0\leq t\leq 60$ with respect to time.

Function ​$F(t)=10+6\cos(\frac{t}{\pi})$

Integrate w.r.t t in the interval $0\leq t\leq 60$  

$\int\limits^{60}_{0} {10+6\cos(\frac{t}{\pi})} \, dt$

$=[10t+6\pi\sin(\frac{t}{\pi})}]^{60}_{0}$

$=10(60)+6\pi\sin(\frac{60}{\pi})-0$

$=600+6.167$

$=606.167$

Approximately, 606 customers arrive at the counter over the 60 minute period.

Answer 2

Answer: 606 costumers ( Approx )

Explanation:

Here, the function that shows the the rate at which customers arrive at a counter,

$F (t)=10+6cos(\frac{t}{\pi})$

For 0 ≤ t ≤ 60 minutes,

The number of costumers,

$\int_{0}^{60} F(T)$

$\int_{0}^{60} 10+6cos(\frac{t}{\pi}) dt$ -----(1)

Let $\frac{t}{\pi}=u$

⇒ $t=\pi u$

⇒ $dt=\pi du$

By substituting this on equation (1),

$=\pi \int_{0}^{60} 10+6 cos u du$

$=\pi [ 10u + 6 sin u ]_{0}^{60}$

$=\pi [ 10 \frac{t}{\pi} + 6 sin (\frac{t}{\pi}) ]_{0}^{60}$

$=\pi [ 10 \frac{60}{\pi} +6 sin (\frac{60}{\pi})- 0]$

$=600+6.16735797767$

$=606.16735797767\approx 606$