maximunm value.

[man111]

Let [tex]f(x,y) = \sqrt{x^2+(y-1)^2}+\sqrt{(x-3)^2+(y-4)^2}-\sqrt{x^2+y^2}-\sqrt{(x-1)^2+y^2}[/tex]. Then Max.[tex]f(x,y)=[/tex],
Where [tex]x,y\in R[/tex]

[martin123456]

Well, I came up with something.
On the coordinate system draw the points [tex](0,1)[/tex], [tex](3,4)[/tex], [tex](0,0)[/tex] and [tex](1,0)[/tex].
Then [tex]\sqrt{(x-a)^2+(y-b)^2}[/tex] is the distance between the points [tex](x,y)[/tex] and [tex](a,b)[/tex].
From the triangle inequality we get that [tex]\sqrt{x^2+(y-1)^2}-\sqrt{(x-1)^2+y^2} \le \sqrt{2}[/tex] and [tex]\sqrt{(x-3)^2+(y-4)^2}-\sqrt{x^2+y^2} \le 5[/tex]. Then it follows that [tex]f(x,y) \le 5+\sqrt{2}[/tex].
When does the equality occurs? Well, let [tex]g[/tex] be the line containing the points [tex](0,0)[/tex] and [tex](3,4)[/tex] and [tex]l[/tex] be the line containing the points [tex](0,1)[/tex] and [tex](1,0)[/tex]. So the equations of these lines are [tex]g: y=\frac{4x}{3}[/tex] and [tex]l:y=-x+1[/tex]. Equality occurs [tex]\Leftrightarrow (x,y) = g\cap l[/tex]. This point is easily calculated and it is [tex](\frac{3}{7}, \frac{4}{7})[/tex].

p.s. Yet, I feel that something is missing - there is another possible rearrangement of the points and the result then is [tex]1+2\sqrt{5}[/tex]. It is smaller than the above but achieved in the same manner. If someone can light me up?

[man111]

Thanks martin u r saying Right ans = [tex]5+\sqrt{2}[/tex]

and I also have a same Confusion like martin.......

can someone explain It.....

[martin123456]

Maybe the reason is that in the first arrangement the intersection point is internal for the segments which is not the case with the second possible arrangement.