设a=2cosα,b=2sinα
4a+3b
=8cosα+6sinα
=2(4cosα+3sinα)
=2(4cosα/5+3sinα/5)*5
=10sin[α+arcsin(4/5)]
注5²=3²+4²,从而构成2个三角函数值
所以最大值为10