一.若x,y满足x+2y-3≤0,x+3y-3≥0,y-1≤0.若目标函数z=ax+y（

1．x+2y-3=0,x+3y-3=0,y-1=0,两两联立解得x1=3,y1=0；x2=1,y2=1；x3=0,y3=1满足约束条件x+2y-3≤0,x+3y-3≥0,y-1≤0的点在由点(3,0),(1,1),(0,1)围成的三角形区域内∵目标函数z=ax+y（a＞0）,在点（3,0）处取最大值Z1=3a,z2=a+1,z3=13a>a+1==>a>1/2,3a>1==>a>1/3∴a∈(1/2,+∞)2．∵y=-sin²a-2msina-2m-1设F(x)=-(sinx)^2-2msinx-2m-1F’(x)=-2(sinx)cosx-2mcosx=-2cosx(sinx+m)=0Cosx=0==>x1=2kπ-π/2,x2=2kπ+π/2,函数f(x)取极小值sinx=-m==>x3=arcsin(-m),x4=π-arcsin(-m) ,函数f(x)取极大值 F(-m)=m^2-2m-11-√2x3=2kπ-arccosa,x4=2kπ+arccosa ,函数f(x)取极小值 f(-arccosa )=-a^2-a,f(arccosa )=-a^2-a当a>1时sinx=0==>x1=2kπ,x2=2kπ+πf”(x)= -2cos(2x)+2acosxf”(0)=2a-2>0,函数f(x)取极小值f(0)＝1-3a；f”(π)=-2a-2x3=2kπ-arccosa,x4=2kπ+arccosa ,函数f(x)取极小值 f(-arccosa )=f(arccosa )=-a^2-a当a>1时,x1=2kπ,函数f(x)取极小值f(0)＝1-3a； x2=2kπ+π,函数f(x)取极大值f(π)＝1+a当a