October 1, 2021
I recently purchased a new Velo-ton stationary bike and took it for a spin. The bike records three key metrics throughout the ride: cadence (how fast I’m riding), resistance (how hard I have to push the pedals to keep moving) and output (the power I produce).
With a little experimentation, I determine that the power (in watts) is equal to the product of the cadence and resistance values divided by 20. For example, if my cadence is 64 and my resistance is 25, then my power output is (64·25)/20, or 80 watts.
Whenever I ride, I always make sure that my resistance is between 20 and 60, while my cadence is between 60 and 100. After a particularly grueling 30-minute workout, I notice that my average resistance was 40, while my average cadence was 80. (Note that these averages are computed per unit of time, rather than per unit of distance traveled.)
At first, I figure my average power was (40·80)/20, or 160 watts. But I soon realize other values are also possible. What is the maximum average power that I could have produced? What is the minimum?
Maximum Average Power: 180 watts
Minimum Average Power: 140 watts
Explanation:
At first this puzzle seemed impossible to solve: don't we require the actual functions for \(c(t)\) and \(r(t)\) (denoting cadence and resistance in terms of units of time in minutes, respectively)?
However, it turns out that with the limits of those functions given, we can figure out the bounds on the average value of the power. The key is to realize that \(c(t)\) and \(r(t)\) need not be in sync for the entire duration of the 30-minute exercise. We can leverage different values of those functions to acquire the average value of \(p(t)\), the power over this time period, in the boundary cases.
Remember:
\begin{align} p(t) &= \frac{c(t)r(t)}{20} \end{align}
For maximum average power, naturally we want to sync up the maximum values of \(c(t)\) and \(r(t)\) for as long as possible. The maximum of \(c(t)\) is given to be 100 while the maximum for \(r(t)\) is given to be 60. Their product yields a whopping \(100 \times 60 = 6000\). We can maintain this for up to 15 minutes. After that, in order to balance out and make sure the average values of those factors are maintained, we need to have \(c(t)\) and \(r(t)\) take on their minimum values, which yields \(60 \times 20 = 1200\). The average power in this case still yields \(p(t) = \frac{6000 + 1200}{2}\frac{1}{20} = 180\) watts, which is actually higher than if we maintained a constant \(c(t)\) and \(r(t)\) (yielding average power of 140 watts in that case).
Is 180 the best we can do? After trying some other schemes, such as linearly increase both \(c(t)\) and \(r(t)\) from their minimum values to their maximum values, the average power in this case is 166.7 (try it!), which is still less than 180. So it seems that 180 is the maximum achievable average power. An argument can be made whereby we "sacrifice" the time allotted for the "6000" product in the 30 minute window, just so that we can increase our minimum product from 1200 to some slightly higher value for some slightly higher duration. The opportunity cost of losing 6000/unit time cannot be effectively made up by raising the minimum to some higher value, as the extreme of this scheme is to keep both \(c(t)\) and \(r(t)\) to their average values constantly, which we know is suboptimal. Therefore, 180 is the best we can do by maximizing the time where "6000" happens (for an entire 15 minutes!).
For the minimum average power, we think the other way. If when we are achieving maximum cadence, let's have minimum resistance, and vice versa. Again, we swap the two pairs half way in the exercise, so effectively we have:
Maximum candence and minimum resistance: \(p = 100 \times 20 = 2000\).
Minimum candence and maximum resistance: \(p = 60 \times 60 = 3600\).
Average power: \(p(t) = \frac{2000 + 3600}{2}\frac{1}{20} = 140\) watts.
Again, an argument about opportunity cost, similar to the maximum average power case, can be made to justify the above value.