Maximum power transfer
The problem as described has a trivial solution.  To maximize the power into the \$100 \Omega\$ load, the pot should be turned to one end so that \$R = 500 \Omega\$.  Then the \$12V\$ source appears entirely across the load and delivers \$\dfrac{12^2}{100}\$ or \$1.44 W\$.  The maximum power theorem applies when the source impedance is fixed and the load is variable, which is not the case here.