(5-9)   ①  Both the examples in the picture give the same answer:

The central peak extends from around -λ(kmax)/2 to +λ(kmax)/2. So the uncertainty in the position of the particle has a magnitude of approximately λ(kmax).

 ②  Since λ= 2π/k, the waveform with k=kmax will have the smallest value of λ.

Then, since p=h/λ, the smallest λ will be associated with the greatest magnitude of the momentum, p.

Hence the greatest momentum is associated with the the waveform that has k=kmax.

③  In turn, this means that the possible momentum of the particle ranges from -h/λ(kmax) to +h/λ(kmax). It could be anywhere in between, arising from the waveforms with k<kmax.  So the uncertainty in the momentum has a magnitude of 2h/λ(kmax).

④   So we have our two uncertainties:

For position:  Δx  =  λ(kmax)

For momentum, Δp = 2h/λ(kmax)

It’s clear that these are inversely proportion, one to the other. So if we multiply them together, we get:

                          ΔxΔp = λ(kmax) x 2h/λ(kmax)  =  2h