Question

求曲线x^3-xy+y^3=1(x》0,y》0)上的点到坐标原点的最长距离与最短距离

Answer 1

做法如下：

x^3-xy+y^3=1(x>0,y>0),

设u=x+y>0,v=xy>0,则u(u^2-3v)-v=1,

u^3-1=(3u+1)v,①

v=(u^3-1)/(3u+1),

由v>0得u>1;

由v<=u^2/4和①，得4(u^3-1)<=(3u+1)u^2,

∴u^3-u^2-4<=0,

(u-2)(u^2+u+2)<=0,u^2+u+2>0,

∴1<u<=2.

曲线x^3-xy+y^3=1(x>0,y>0)上的点到坐标原点的距离

d=√(x^2+y^2)=√(u^2-2v)

=√[u^2-2(u^3-1)/(3u+1)]

=√[(u^3+u^2+2)/(3u+1)],

w=d^2=(u^3+u^2+2)/(3u+1)(u>1),

w'=[(3u^2+2u)(3u+1)-3(u^3+u^2+2)]/(3u+1)^2

=[9u^3+9u^2+2u

 -3u^3-3u^2    -6]/(3u+1)^2

=(6u^3+6u^2+2u-6)/(3u+1)^2>0,

∴w是u的增函数，w(1)=1,w(2)=2，

∴d无最小值，有最大值√2，下确界1。

Answer 2

简单计算一下即可，答案如图所示

Answer 3

x^3-xy+y^3=1(x>0,y>0),
设u=x+y>0,v=xy>0,则u(u^2-3v)-v=1,
u^3-1=(3u+1)v,①
v=(u^3-1)/(3u+1),
由v>0得u>1;
由v<=u^2/4和①，得4(u^3-1)<=(3u+1)u^2,
∴u^3-u^2-4<=0,
(u-2)(u^2+u+2)<=0,u^2+u+2>0,
∴1<u<=2.
曲线x^3-xy+y^3=1(x>0,y>0)上的点到坐标原点的距离
d=√(x^2+y^2)=√(u^2-2v)
=√[u^2-2(u^3-1)/(3u+1)]
=√[(u^3+u^2+2)/(3u+1)],
w=d^2=(u^3+u^2+2)/(3u+1)(u>1),
w'=[(3u^2+2u)(3u+1)-3(u^3+u^2+2)]/(3u+1)^2
=[9u^3+9u^2+2u
-3u^3-3u^2 -6]/(3u+1)^2
=(6u^3+6u^2+2u-6)/(3u+1)^2>0,
∴w是u的增函数，w(1)=1,w(2)=2，
∴d无最小值，有最大值√2，下确界1。