What is the total cost to heat the greenhouse during this 24-hour period? The temperature outside the greenhouse during a $24$-hour period is given by $F(t) = 22 + 20\cos(\frac{\pi t}{12}), 0 \le t \le 24$, where $F(t)$ is measured in degrees Fahrenheit and t is measured in hours. Let $t=0$ represent noon on the first day.
d. The heater in the greenhouse kicks on whenever the outdoor temperature is at or below $36$ degrees Fahrenheit. For what values of $t$ is the heater on?
I used a graphing calculator for this part, between $[3.082, 20.962]$.
e. The cost of heating the greenhouse accumulates at a rate of $\$0.06$ per hour for each degree the outside temperature falls below $36$ degrees Fahrenheit. What is the total cost to heat the greenhouse during this $24$-hour period? 
I'm not really understanding where to start on this problem. My teacher suggested finding the integral difference between $y=36$ and $F(t)$ when the degrees fall under $36$, but I'm confused on what this represents and what to do with it. If anyone could help me understand, that would be much appreciated.
Thank you!
 A: The temperature outside is $$F(t)=22+20\cos(\frac{\pi t}{12})$$ The temperature difference if $\Delta T(t)=36-F(t)$, but you only need to consider it if $F(t)\lt 36$, or $\Delta T>0$. Let's consider a small interval of time $d t$. Since the temperature difference is $\Delta T$, the price for this interval is $$0.06\Delta T(t) dt$$ You need to add together the prices for all these small time intervals, which means doing the integral.
$$\mathrm{Cost}=\int_{t_1}^{t_2}0.06(36-22-20\cos(\frac{\pi t}{12}))dt$$
Here $t_1$ is when the temperature dips below 36 degrees, and $t_2$ is when it raises to 36 again. 
A: This question is similar with question 5 from the 1998 Calc AB FRQ . Only the numbers are changed. 
So you have found $$t\in[3.082,20.962]$$
Then the setup for part e. will be the following:
$$\begin{align}
C_{total~cost} &= 0.06\int_{3.082}^{20.962}\left(36-\left[22+20\cos\left(\frac{\pi t}{12}\right)\right]\right)dt\\
&=0.06(360.03979)=21.602\approx$21.6
\end{align}$$
A: Just for the fun of it.
You used a  graphing calculator for solving
$$22 + 20\cos(\frac{\pi t}{12})=36 \implies \cos(\frac{\pi t}{12})=\frac 7{10}$$ You could have used the approximation
$$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad (-\frac \pi 2 \leq x\leq\frac \pi 2)$$ making $x=\frac{\pi t}{12}$, we end then with
$$\frac{144-4 t^2 }{144+t^2}=\frac 7{10}\implies 47t^2=432\implies t=12 \sqrt{\frac{3}{47}}\approx 3.03175$$
$$C_{total~cost} = \frac 6 {100} \int_{12 \sqrt{\frac{3}{47}}}^{24-12 \sqrt{\frac{3}{47}}}\left(36-\left[22+20\cos\left(\frac{\pi t}{12}\right)\right]\right)dt$$ making the nice
$$C_{total~cost}=\frac{504 \left(47-\sqrt{141}\right)}{1175}+\frac{144 }{5 \pi }\sin \left(\sqrt{\frac{3}{47}} \pi \right)$$ We can continue using the magnificent approximation
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$  proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (that is to say more than $1400$ years ago)
$$\sin \left(\sqrt{\frac{3}{47}} \pi \right)\simeq \frac{16 \left(235 \sqrt{141}-177\right)}{58753}$$
$$C_{total~cost}=\frac{504 \left(47-\sqrt{141}\right)}{1175}+\frac{2304 \left(235 \sqrt{141}-177\right)}{293765 \pi }\approx 21.5912$$ while a completely rigorous calculation would give
$$C_{total~cost}=\frac{72 \left(\sqrt{51}+7 \pi -7 \cos ^{-1}\left(\frac{7}{10}\right)\right)}{25
   \pi }\approx 21.6026$$
I hope and wish that we shall not argue for one cent difference.
Merry Xmas
