a . 14； b . 16； c . 17； D. 19
Solution: make a straight line l? : x+2y-5=0, let its intersection with the X axis be a (5 5,0); Make another straight line? : 2x+y-7=0, let it be equal to l? about
The intersection (3, 1) is b, and the intersection (0, 7) with the y axis is c; Then the region defined by the inequality group {x+2y-5 > 0, 2x+y-7 > 0, x≥0 y≥0} is a semi-open region surrounded by the upper half of the X axis (including the X axis), the right half of the Y axis (including the Y axis) and the upper half of the polyline ABC.
Because the inequalities X+2Y-5 > 0 and 2x+Y-7 > 0 have no equal sign, none of the points on the dotted line ABC can be counted in the area specified above.
Inside. And both x and y are integers, then the points closest to the boundary of this region are arranged from right to left as follows: (6,0); (5, 1); (4, 1)
(3,2); (2,4); ( 1,6); (0,8).* * * 7 points, then the point that minimizes the value of 3x+4y among these points is point (4, 1), and its value =3×4+4× 1= 16, so we should choose B.