Brilli's Diet Pill Desperation

The torque due to Brilli's weight is $\tau = Mgx = M_0e^{-\alpha t}gx = M_0gxe^{-\frac{\alpha}v x}.$ It varies with position $x$ according to $\frac{d\tau}{dx} = M_0ge^{-\frac{\alpha}v x}\left(1 - \frac{\alpha}vx\right).$ At the maximum, this derivative is zero. Thus $0 = \frac{d\tau}{dx} = 1 - \frac{\alpha}vx;$ $x = \frac v \alpha = \frac{\SI{0.02}{m/s}}{\SI{0.001}{s^{-1}}} = \boxed{\SI{20}{m}}.$

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Incidentally, this is a good example to give an intuitive interpretation of the use of the product rule.

The torque $Mgx$ due to Brilli's weight undergoes to kinds of change. For every kilogram lost by Brilli, it decreases by $gx$ . For every meter walked, it increases by $Mg$ . At first, the increase is stronger than the decrease. Later it is the other way around. The maximum is reached when these two factors balance: $gx\cdot (\text{rate of weight loss}) = Mg\cdot (\text{rate of walking}).$

Now Brilli loses weight at a rate $\alpha M$ and gains distance at a rate $v$ . Thus $gx\cdot \alpha M = Mg\cdot v.$ This leads to the same conclusion as before, namely $x = v/\alpha$ .