c
\(\begin{array}{l}
{\rm{Minimum}}\,{\rm{force}}\,{\rm{on}}\,{\rm{A}}\\
{\rm{ = 12}}\,{\rm{N}}\\
{\rm{So,}}\,{\rm{both}}\,{\rm{block}}\,{\rm{to}}\,{\rm{move}}\,{\rm{together}}\\
12 - \mu  \times 4 \times 10 = 40 \times \frac{{12}}{9}\\
 \Rightarrow \mu \, = \frac{1}{6}
\end{array}\)

\(\begin{array}{l}
Now,\,{\rm{maximum}}\,force\,F\,applied\,on\\
lower\,block\,so,\,as\,to\,move\,together\\
F - \mu \,4 \times 10 = 5 \times \left( {\frac{{\mu  \times 4 \times 10}}{4}} \right)\\
 \Rightarrow \,F = \mu  \times 10\left( {5 + 4} \right)\\
 = \frac{1}{6} \times 10 \times 9 = 15\,N
\end{array}\)