已知实数xy满足1/2x^2+2x+y-1=0,则x+2y的最大值为
问题描述:
已知实数xy满足1/2x^2+2x+y-1=0,则x+2y的最大值为
答
½x²+2x+y-1=0
y=-½x² -2x+1
x+2y=x-x²-4x+2=-x²-3x+2=-(x+ 3/2)²+17/4
当x=-3/2时,x+2y有最大值(x+2y)max=17/4