（1）
易知直线与x轴交点为 (-(b+2)/k,0),与y轴交点为(0,b+2)
因交于正半轴,则 -(b+2)/k >0,b+2>0,k0 ,b+5>0 ,k 0
而
S = |OA|+|OB|+3 = |-(b+2)/k| + |b+2| + 3 = -(b+2)/k + (b+2) +3
将k = -b(b+2)/2(b+5) 代入,得
S = 2(b+5)/b + (b+2) +3 = 10/b + b + 7
因b>0
由基本不等式得
S = 10/b + b + 7 ≥ 2√[(10/b)*b] + 7 = 7 + 2√10
当且仅当 10/b = b,即 b = √10时
△OAB面积取得最小值为 Smin = 7 + 2√10