(A) What is the temperature of the greenhouse at 1:00 a.m. to the nearest Fahrenheit?

(B) Find the average temperature between 7:00 p.m. and 7:00 a.m. to the nearest tenth of a degree Fahrenheit.

(C) When the temperature of the greenhouse drops below 80 degreeso Fahrenheit, a heating system will automatically be turned on a maintain the temperature at minimum of 80 degrees Fahrenheit. At what values of the to the nearest tenth is the heating system turned on?

(D) The cost of heating the greenhouse is $0.25 per hour for each degree. What is the total cost to the nearest dollar to heat the greenhouse from 7:00 p.m. and 7:00 a.m.?
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The temperature in a greenhouse from 7:00p.m. to 7:00a.m. is given by f (t)= 96 - 20sin (t/4), where f (t) is measured in Fahrenheit, and t is the number of hours since 7:00 p.m.
A) What is the temperature of the greenhouse at 1:00 a.m. to the nearest Fahrenheit?

(B) Find the average temperature between 7:00 p.m. and 7:00 a.m. to the nearest tenth of a degree Fahrenheit.

(C) When the temperature of the greenhouse drops below 80 degreeso Fahrenheit, a heating system will automatically be turned on a maintain the temperature at minimum of 80 degrees Fahrenheit. At what values of the to the nearest tenth is the heating system turned on?

(D) The cost of heating the greenhouse is $0.25 per hour for each degree. What is the total cost to the nearest dollar to heat the greenhouse from 7:00 p.m. and 7:00 a.m.?

 


The equation is: f(t) = 96 – 20sin(t/4), 0 <= t <= 12


(A)  At 1.00 a.m. t = 6
f(6) = 96 – 20.sin(6/4) = 96 – 20*0.99749 = 96 – 19.9499
f(6) = 76 ⁰F

(B)  The average temperature would need to be worked out by sampling the temperature at different times throughout the night.
Divide the temperature range into N equal intervals, giving N+1 sampling points.
We would then have T1 = f(δt), T2 = f(2δt), T3 = f(3δt), ... ,Tn = f(nδt)
Where δt = range/N = 12/N, and n = 0..N
Giving Tn = 96 – 20.sin((12n/N)/4) = 96 – 20.sin(3n/N)

Then Tav = (1/N)*sum(Tn, n = 0 .. N)
i.e. Tav = (1/N)*sum(96 – 20.sin(3n/N), n = 0 .. N
Tav = 96 – 20. (1/N)*sum(sin(3n/N), n = 0 .. N


I used Maple to evaluate the above summation.
The results are tabulated as follows.

                                 Average Temperature
Num Intervals            3       4        6       10       20      50     100    200
Tav over the range 83.39 83.01 82.78 82.69 82.69 82.71 82.72 82.73

As can be seen from the table the temperature is averaging out at:
 Tav = 82.7 ⁰F

(C)  T = f(t) = 96 – 20sin(t/4), 0 <= t <= 12
At T = 80 ⁰F,            96 – 20sin(t/4) = 80
20.sin(t/4) = 96 – 80 = 16
sin(t/4) = 0.8
t/4 = 0.927295
t = 3.70918
t = 3.7 (to nearest tenth)

(D)  The temperature will (normally) drop to 80 ⁰F after t = 3.7 hours and rise again to 80 ⁰F when t = 12 – 3.7 = 8.3 hours.
Heating system is turned on for 8.3 – 3.7 = 4.6 hours
Cost of heating is 4.6*80*0.25 = 4.6*20 = 92
Cost = $92

 

by Level 11 User (81.5k points)

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