设实数xyz满足x+y+2z=4 xy+3yz+3zx=7 求z的最大值
问题描述:
设实数xyz满足x+y+2z=4 xy+3yz+3zx=7 求z的最大值
答
x+y=4-2zxy+3z(x+y)=7∴ xy=6z^2-12z+7∵ (x+y)^2≥4xy∴ (4-2z)^2≥4(6z^2-12z+7)∴ 16-16z+4z^2≥4(6z^2-12z+7)∴ 5z^2-8z+3≤0即 (z-1)(5z-3)≤0解得,3/5≤z≤1z的最大值为1...