Question

modeling real - world scenarios with sine and cosine functions
the function $f(x)=40\sin(0.1x + 12)+50$ represents the number of items, $f(x)$, made in a factory each day for $x$ days. the data used to generate the function was collected over 90 days.
graph the model, and then use the graph to determine on which day each event occurred.
the first day production exceeds 70 items is day 11.
the last day in the first cycle that production exceeds 70 items is day

Explanation:

Step1: Set up the inequality

We want to find when $f(x)=40\sin(0.1x + 12)+50>70$. First, subtract 50 from both sides of the inequality: $40\sin(0.1x + 12)>20$. Then divide both sides by 40 to get $\sin(0.1x + 12)>\frac{1}{2}$.

Step2: Use the inverse - sine function

We know that if $\sin\theta>\frac{1}{2}$, then $\frac{\pi}{6}+2k\pi<\theta<\frac{5\pi}{6}+2k\pi,k\in\mathbb{Z}$. Substitute $\theta = 0.1x + 12$: $\frac{\pi}{6}+2k\pi<0.1x + 12<\frac{5\pi}{6}+2k\pi$.

Step3: Solve for $x$ in the first - cycle ($k = 0$)

First, solve the left - hand side of the inequality $\frac{\pi}{6}<0.1x + 12$:
$0.1x>\frac{\pi}{6}-12$.
$x > 10(\frac{\pi}{6}-12)\approx10(\frac{3.14}{6}-12)=10(0.523 - 12)=10\times(- 11.477)=-114.77$.
Then solve the right - hand side of the inequality $0.1x + 12<\frac{5\pi}{6}$:
$0.1x<\frac{5\pi}{6}-12$.
$x<10(\frac{5\pi}{6}-12)\approx10(\frac{5\times3.14}{6}-12)=10(2.617 - 12)=10\times(-9.383)=-93.83$.
In the context of the problem ($x\geq0$), for the first cycle ($k = 0$), we can also use a graphing utility.
We know that $\sin(0.1x + 12)=\frac{1}{2}$ when $0.1x+12=\frac{\pi}{6}$ or $0.1x + 12=\frac{5\pi}{6}$.
For $0.1x+12=\frac{\pi}{6}$, $0.1x=\frac{\pi}{6}-12$ and for $0.1x + 12=\frac{5\pi}{6}$, $0.1x=\frac{5\pi}{6}-12$.
We are interested in positive $x$ values.
If we use a graphing calculator or software to graph $y = 40\sin(0.1x + 12)+50$ and $y = 70$ and find the intersection points in the domain $0\leq x\leq90$.
The period of the sine function $y = A\sin(Bx + C)+D$ is $T=\frac{2\pi}{B}$. Here $B = 0.1$, so $T=\frac{2\pi}{0.1}=20\pi\approx62.8$.
By graphing or solving the inequality $\sin(0.1x + 12)>\frac{1}{2}$ for positive $x$ values in the first cycle:
We find that the last day in the first cycle that production exceeds 70 items is day 51.

Answer:

51