Answer :
The temperature exceeds 76°F for the entire 24-hour period.
The total cost of cooling the house for the month of June is $57.60.
a. To find the average temperature over the 24-hour period, we need to integrate the function F(t) over the interval [0, 24] and then divide the result by the length of the interval.
The integral of 11cos(12π(t-15)) from 0 to 24 can be found by using the formula for the integral of cosine function: ∫cos(x)dx = sin(x) + C.
Let's plug in the limits of integration and evaluate the integral:
∫[0,24] 11cos(12π(t-15)) dt
= 11∫[0,24] cos(12π(t-15)) dt
= 11[ sin(12π(t-15))/(12π) ] evaluated from 0 to 24
= 11[ sin(12π(24-15))/(12π) - sin(12π(0-15))/(12π) ]
= 11[ sin(108π)/(12π) - sin(-180π)/(12π) ]
= 11[ sin(108π)/(12π) + sin(180π)/(12π) ]
Now, the average temperature is obtained by dividing the integral by the length of the interval:
Average temperature = (1/24) * 11[ sin(108π)/(12π) + sin(180π)/(12π) ]
b. To find the interval of time when the air conditioner will be cooling the house, we need to determine when the outside temperature is above 76°F. We can achieve this by solving the inequality F(t) > 76:
81 + 11cos(12π(t-15)) > 76
Simplifying this inequality leads to:
11cos(12π(t-15)) > -5
To solve this inequality, we need to consider the cosine function and its range. Since the maximum value of the cosine function is 1, we have:
11cos(12π(t-15)) > -5
cos(12π(t-15)) > -5/11
The cosine function is greater than -5/11 for all values of t. Therefore, the air conditioner will be cooling the house for the entire 24-hour period.
c. To find the total cost of cooling the house for the month of June, we need to determine the number of hours the temperature exceeds 76°F for each day and multiply it by the rate of 8 cents per hour. Then, we sum up these costs for all 30 days.
Let's calculate the total cost for one day first. We need to find the number of hours the temperature exceeds 76°F. This can be done by setting F(t) > 76 and solving for t. Using the same inequality from part b:
11cos(12π(t-15)) > -5
We can solve this inequality by considering the cosine function and its range. Since the maximum value of the cosine function is 1, we have:
11cos(12π(t-15)) > -5
cos(12π(t-15)) > -5/11
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