/sci/ - Science & Math

>>12759816
You should probably understand the maximization in this problem, since it's pretty simple. I think it's pretty clear how maximizing our denominator results in a minimum for P. Since our denominator [math] \cos(\phi) -\mu_k \sin(\phi) [/math] is a continuous & differentiable function of phi, by interior extremum theorem we know that it must attain its maxima and minima when its derivative is equal to 0, so we just find the phi for which that condition is met and check where it attains a maximum or minimum there.

If you haven't seen this theorem before and want a quick intuition for it, suppose that some continuous & differentiable [math] f(x) [/math] achieves its maximum at some [math] x_0, [/math] and [math] f'(x_0) > 0 [/math]. Then if you take a step of length delta to the right that's small enough, [math] f(x_0+\delta) \approx f(x_0) + \delta f'(x_0) > f(x_0), [/math] so [math] x_0 [/math] can't be a maximum. The same style of argument hold when [math] f'(x_0) < 0 [/math] by taking a step to the left instead of the right, and so if [math] x_0 [/math] is a maximum, [math] f'(x_0) = 0. [/math]