MATH SOLVE

4 months ago

Q:
# 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.

Accepted Solution

A:

Answer:[tex]f(x)=8500\sin\left(\frac{\pi}{6}\left(x-1\right)\right)+33500[/tex]Step-by-step explanation:Given information:Maximum revenue = 42000Minimum revenue = $25000Time period = 12 monthsWe need to write a function that models the boutique's revenue during the year, where corresponds to the month.[tex]f(x)=Asin(B[x-C])+D[/tex] .... (1)where, A is amplitude, [tex]\frac{2\pi}{B}[/tex] is period, C is phase shift and D is midline.[tex]A=Amplitude =\frac{Maximum-Minimum}{2}\Rightarrow \frac{42000-25000}{2}=8500[/tex][tex]D=midline =\frac{Maximum+Minimum}{2}\Rightarrow \frac{42000+25000}{2}=33500[/tex][tex]Period=\frac{2\pi}{B}[/tex][tex]12=\frac{2\pi}{B}\Rightarrow B=\frac{\pi}{6}[/tex]Substitute the value of A, B and D in equation (1).[tex]f(x)=8500\sin\left(\frac{\pi}{6}\left(x-C\right)\right)+33500[/tex] ..... (2)In April, revenue of the store is $42000. So, the graph passes through the point (4,42000).[tex]42000=8500\sin\left(\frac{\pi}{6}\left(4-C\right)\right)+33500[/tex][tex]42000-33500=8500\sin\left(\frac{\pi}{6}\left(4-C\right)\right)[/tex][tex]8500=8500\sin\left(\frac{\pi}{6}\left(4-C\right)\right)[/tex]Divide both sides by 8500.[tex]1=\sin\left(\frac{\pi}{6}\left(4-C\right)\right)[/tex][tex]\sin \frac{\pi}{2}=\sin\left(\frac{\pi}{6}\left(4-C\right)\right)[/tex]On comparing both sides we get[tex]\frac{\pi}{2}=\frac{\pi}{6}(4-C)[/tex][tex]3=4-C[/tex][tex]C=4-3[/tex][tex]C=1[/tex]Substitute the value of C in equation (2).Therefore, the required function is [tex]f(x)=8500\sin\left(\frac{\pi}{6}\left(x-1\right)\right)+33500[/tex].