A boy of mass M exerts a horizontal force to slide a box of mass M'...

The maximum horizontal force the boy can exert without slipping backward is given by the limiting friction between his shoes and the floor, F_\max = \mu M g.

The minimum horizontal force required to slide the box is the limiting friction between the box and the floor, f_\max = \mu' M' g.

For the boy to slide the box, he must be able to apply a force greater than the friction on the box without slipping himself. The condition to slide the box is F_\max > f_\max, which gives $\mu M g > \mu' M' g$, or $\mu M > \mu' M'$.

It is definitively impossible to slide the box if the maximum force the boy can exert is less than the force required to slide the box, which means $\mu M < \mu' M'$.

Analyzing the given options, if $\mu < \mu'$ and $M < M'$, then multiplying these two inequalities for positive quantities yields $\mu M < \mu' M'$. Under this condition, the boy can never exert enough force to slide the box.