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Math  /  Algebra

QuestionAn air conditioner cools a home when the outside temperature is about 78 EdegF. During the summer you can model the outside temperature in degrees Fahrenheit using the function f(t)=72−9cos⁡(112πt)f(t)=72-9 \cos \left(\frac{1}{12} \pi t\right) where tt is the number of hours past midnight. During what hours is the air conditioner cooling the home? Round the hour to nearest tenth.
Time format is a 24 hour clock, please pay attention to notes beside answer boxes (for auto-grading purposes).
The air conditioner comes on approximately □\square hours after midnight or □\square A.M. (enter answer as h:mm *do not put a zero in front of the hour for answers like: 3:25, do NOT enter as 03:25)
By the symmetry of the graph, it goes off about □\square hours before midnight or □\square P.M. (enter answer as hh:mm *for an answer like 7:15 pm enter using 24 hour format as: 19:15)

Studdy Solution

STEP 1

1. The air conditioner cools the home when the outside temperature is about 78°F.
2. The function f(t)=72−9cos⁡(112πt) f(t) = 72 - 9 \cos \left(\frac{1}{12} \pi t \right) models the outside temperature in degrees Fahrenheit.
3. t t is the number of hours past midnight.
4. We need to find the time intervals when f(t)≥78 f(t) \geq 78 .

STEP 2

1. Set up the inequality for when the air conditioner is cooling.
2. Solve the inequality for t t .
3. Interpret the solution in terms of time.

STEP 3

Set up the inequality for when the air conditioner is cooling:
f(t)≥78 f(t) \geq 78
Substitute the given function:
72−9cos⁡(112πt)≥78 72 - 9 \cos \left(\frac{1}{12} \pi t \right) \geq 78

STEP 4

Solve the inequality:
1. Rearrange the inequality:
−9cos⁡(112πt)≥6 -9 \cos \left(\frac{1}{12} \pi t \right) \geq 6
2. Divide by −9-9 (note the inequality sign changes direction):
cos⁡(112πt)≤−23 \cos \left(\frac{1}{12} \pi t \right) \leq -\frac{2}{3}
3. Find the general solution for cos⁡θ=−23 \cos \theta = -\frac{2}{3} :
θ=cos⁡−1(−23) \theta = \cos^{-1} \left(-\frac{2}{3}\right)
4. Calculate the specific angles:
θ≈2.3005 \theta \approx 2.3005 radians and \( \theta \approx 2\pi - 2.3005 \approx 3.9827 \] radians

STEP 5

Convert the angles to time:
1. Use the relationship θ=112πt \theta = \frac{1}{12} \pi t :
112πt=2.3005 \frac{1}{12} \pi t = 2.3005
t=2.3005×12π≈8.8 t = \frac{2.3005 \times 12}{\pi} \approx 8.8
2. For the second angle:
112πt=3.9827 \frac{1}{12} \pi t = 3.9827
t=3.9827×12π≈15.2 t = \frac{3.9827 \times 12}{\pi} \approx 15.2

STEP 6

Interpret the solution in terms of time:
1. The air conditioner comes on approximately 8.8 hours after midnight, which is:
8:48 A.M. 8:48 \, \text{A.M.}
2. By symmetry, it goes off approximately 15.2 hours after midnight, which is:
15:12 P.M. 15:12 \, \text{P.M.}
The air conditioner comes on approximately 8.8 8.8 hours after midnight or 8:48 A.M. 8:48 \, \text{A.M.} .
By the symmetry of the graph, it goes off about 15.2 15.2 hours after midnight or 15:12 P.M. 15:12 \, \text{P.M.} .

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