Force needed to hold a book between your hands - not sure about the influence of friction

I'm looking at this problem:

A student presses a book between his hands, as the drawing indicates. The forces that he exerts on the front and back covers of the book are perpendicular to the book and are horizontal. The book weighs $31 \:\mathrm{N}$. The coefficient of static friction between his hands and the book is $0.40$. To keep the book from falling, what is the magnitude of the minimum pressing force that each hand must exert?

The solution given is:

If the minimum pressing force is used, then the frictional force on each cover is equal to the maximum frictional force $\mu_sF_N$ . The weight of the book must be equal to the sum of the two frictional forces: $$31 = 2\mu_sF_N = 2(0.40)F_N$$

I'm confused though as to why you need to press with enough force to get the maximum amount of static friction in order to minimize pressing force. Couldn't you press slightly less, get say 75% of the static friction, and still hold the book up if that force is greater than its weight?

I'm confused though as to why you need to press with enough force to get the maximum amount of static friction in order to minimize pressing force

If you push on each side of the book with force $F_N$, then what is holding the book? The static friction of course.

Formula for static friction: $F_{s}\leq\mu_s F_N$

Since it works on both sides of the book, two times this friction will hold up the book's weight: $2 F_s=31\:\mathrm{N}$

So far so good.

But now, think about what force is actually needed to hold that book. If you reduce your pressing force $F_N$ on the book, then you also reduce the maximum force the static friction $F_s$ can reach. The actual friction is constant, though, because of $2 F_s=31\:\mathrm{N}$, since the weight to be held doesn't change. But the maximum that the friction can reach now gets closer to the value it must take as you ease on your pressing force.

At some point you have reduced $F_N$ so much that the maximum limit for $F_s$ is equal to the needed value to hold the book. Now the friction is at its maximum, and this corresponds to the minimum pressure you must provide to avoid the book from falling.

The frictional force is the only force in this situation that opposes the weight of the book. The hands in this case are providing the normal force that you will recognize from an inclined plane problem. The friction force is determined from the normal force (that the hands are providing) multiplied by the static coefficient of friction. The friction force must equal the weight to prevent the book from falling.