Question 176633
First find the feasible region and vertices.
{{{1 <= x <= 6}}}
{{{2 <= y <= 5}}}
The last one involves a little more work.
{{{x+y<=10}}}
{{{y<=-x+10}}}
Let's graph the line and the other boundaries
{{{drawing( 300, 300, -1, 11, -1, 11,grid( 1 ),circle( 0, 0, .2 ),
green(line(-2,2,12,2)),
green(line(-2,5,12,5)),
green(line(1,-1,1,15)),
green(line(6,-1,6,15)),
graph( 300, 300, -1, 11, -1, 11, -x+10)) }}}
The feasible region is then, 
{{{drawing( 300, 300, -1, 7, -1, 7,grid( 1 ),
circle( 1, 2, .2 ),
circle( 1, 5, .2 ),
circle( 6, 2, .2 ),
circle( 6, 4, .2 ),
circle( 5, 5, .2 ),

green(line(1,2,6,2)),
green(line(1,2,1,5)),
green(line(5,5,6,4)),
green(line(6,2,6,4)),
green(line(6,4,5,5)),
green(line(5,5,1,5))
) }}}
The vertices are then
(1,2)
(1,5)
(6,2)
(6,4)
(5,5)
The corresponding values of P are
{{{P(1,2)=3(1) + 2(2)=7}}} 
{{{P(1,5)=3(1) + 2(5)=13}}}
{{{P(6,2)=3(6) + 2(2)=22}}}
{{{P(6,4)=3(6) + 2(4)=26}}}
{{{P(5,5)=3(5) + 2(5)=22}}}
The minimum value for P of 7 occurs at (1,2).
The correct answer is A.