1. CS 121 – Quiz 1
Potato Problem

2. The equation for the temperature of the potato takes on the form T = a − b × ct, where a, b, and c are known constants, and t is the time that the potato is in the oven, measured in minutes.
If we know the potato is considered done when its temperature is anywhere between X and Y degrees, we can find the time interval at which the potato will be done by solving the equations X = a − b × ct and Y = a − b × ct for t.

3. We can find the potato’s starting temperature and the oven’s temperature by looking at the equation (T = a − b × ct) itself.
Getting the potato’s starting temperature is easy. It’s the value of T when we first put the potato in the oven (t = 0).
Getting the oven’s temperature is similar. We know that the potato can only get as hot as the oven itself, so we can find the potato’s maximum temperature (and therefore the oven’s temperature) by taking the limit of T as t goes to infinity.

4. Now that we have the potato’s starting temperature and the oven’s temperature (let’s call them P and O), we can find the temperature halfway between them (let’s call this H):
H = (P + O) / 2
We can then plug in H for T and solve the original equation for t (H = a − b × ct)