Question

Over the past several years, the owner of a boutique on Aspen Avenue has observed a pattern in the amount of revenue for the store. The revenue reaches a maximum of about $ 42000 in April and a minimum of about $ 25000 in October. Suppose the months are numbered 1 through 12, and write a function of the form f(x)=Asin(B[x-C])+D that models the boutique's revenue during the year, where corresponds to the month.

Answer 1

Answer:

$f(x)=8500\sin\left(\frac{\pi}{6}\left(x-1\right)\right)+33500$

Step-by-step explanation:

Given information:

Maximum revenue = 42000

Minimum revenue = $25000

Time period = 12 months

We need to write a function that models the boutique's revenue during the year, where corresponds to the month.

$f(x)=Asin(B[x-C])+D$           .... (1)

where, A is amplitude, $\frac{2\pi}{B}$ is period, C is phase shift and D is midline.

$A=Amplitude =\frac{Maximum-Minimum}{2}\Rightarrow \frac{42000-25000}{2}=8500$

$D=midline =\frac{Maximum+Minimum}{2}\Rightarrow \frac{42000+25000}{2}=33500$

$Period=\frac{2\pi}{B}$

$12=\frac{2\pi}{B}\Rightarrow B=\frac{\pi}{6}$

Substitute the value of A, B and D in equation (1).

$f(x)=8500\sin\left(\frac{\pi}{6}\left(x-C\right)\right)+33500$         ..... (2)

In April, revenue of the store is $42000. So, the graph passes through the point (4,42000).

$42000=8500\sin\left(\frac{\pi}{6}\left(4-C\right)\right)+33500$

$42000-33500=8500\sin\left(\frac{\pi}{6}\left(4-C\right)\right)$

$8500=8500\sin\left(\frac{\pi}{6}\left(4-C\right)\right)$

Divide both sides by 8500.

$1=\sin\left(\frac{\pi}{6}\left(4-C\right)\right)$

$\sin \frac{\pi}{2}=\sin\left(\frac{\pi}{6}\left(4-C\right)\right)$

On comparing both sides we get

$\frac{\pi}{2}=\frac{\pi}{6}(4-C)$

$3=4-C$

$C=4-3$

$C=1$

Substitute the value of C in equation (2).

Therefore, the required function is $f(x)=8500\sin\left(\frac{\pi}{6}\left(x-1\right)\right)+33500$.