2x2+3y2=12,X^2/6+Y^2/4=1,令,X=√6*cosa,y=2*sina.x+2y=√6*cosa+2sina=√10*[2/√10*sina+√6/√10*cosa]=√10*sin(a+p),(其中,cosP=2/√10,sinP=√6/√10)则,当sin(a+p)=1时,X+2Y最大值=√10.当sin(a+p)=-1时,X+2...