Question

若实数x,y满足x²+y²-2x+4y=0,则x-2y的最大值为？

Answer 1

(1)
设x-2y=t,代入原式整理得
5y^2+4ty+t^2-2t=0
判别式不小于0,故
△=(4t)^2-4×5×(t^2-2t)≥0
→t^2-10t≤0
→0≤t≤10
故所求最大值为10,最小值为0.
(2)
将原式配方:(x-1)^2+(y+2)^2=5.
依Cauchy不等式,得
[1^2+(-2)^2][(x-1)^2+(y+2)^2]≥[1×(x-1)+(-2)×(y+2)]^2
→5×5≥(x-2y-5)^2
→-5≤x-2y-5≤5
→0≤x-2y≤10.
故所求最大值为10,最小值为0.

Answer 2

x²+y²-2x+4y=0

∴(x-1)²+(y+2)²=5

用圆的参数方程
x=1+√5cost,y=-2+√5sint

x-2y=(1+√5cost)-2(-2+√5sint)=5+√5cost-2√5sint=5+5((√5/5)cost-(2√5/5)sint)

设sina=√5/5,cosa=2√5/5

∴x-2y=5+5(sinacost-sintcosa)=5+5sin(a-t)

∵-1<=sin(a-t)<=1

∴0<=5+5sin(a-t)<=10,即0<=x-2y<=10

最大值为10