Answer: 606 costumers ( Approx )
Explanation:
Here, the function that shows the the rate at which customers arrive at a counter,
[tex]F (t)=10+6cos(\frac{t}{\pi})[/tex]
For 0 ≤ t ≤ 60 minutes,
The number of costumers,
[tex]\int_{0}^{60} F(T)[/tex]
[tex]\int_{0}^{60} 10+6cos(\frac{t}{\pi}) dt[/tex] -----(1)
Let [tex]\frac{t}{\pi}=u[/tex]
⇒ [tex]t=\pi u[/tex]
⇒ [tex]dt=\pi du[/tex]
By substituting this on equation (1),
[tex]=\pi \int_{0}^{60} 10+6 cos u du[/tex]
[tex]=\pi [ 10u + 6 sin u ]_{0}^{60}[/tex]
[tex]=\pi [ 10 \frac{t}{\pi} + 6 sin (\frac{t}{\pi}) ]_{0}^{60}[/tex]
[tex]=\pi [ 10 \frac{60}{\pi} +6 sin (\frac{60}{\pi})- 0][/tex]
[tex]=600+6.16735797767[/tex]
[tex]=606.16735797767\approx 606[/tex]
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 [tex]F(t)=10+6\cos(\frac{t}{\pi})[/tex] [tex]0\leq t\leq 60[/tex] 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 [tex]0\leq t\leq 60[/tex] with respect to time.
Function [tex]F(t)=10+6\cos(\frac{t}{\pi})[/tex]
Integrate w.r.t t in the interval [tex]0\leq t\leq 60[/tex]
[tex] \int\limits^{60}_{0} {10+6\cos(\frac{t}{\pi})} \, dt[/tex]
[tex]=[10t+6\pi\sin(\frac{t}{\pi})}]^{60}_{0}[/tex]
[tex]=10(60)+6\pi\sin(\frac{60}{\pi})-0[/tex]
[tex]=600+6.167[/tex]
[tex]=606.167[/tex]
Approximately, 606 customers arrive at the counter over the 60 minute period.
