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 Three numbers (Posted on 2003-11-17)
If x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3, what is the largest value that x can have ?

 No Solution Yet Submitted by Ravi Raja Rating: 3.7500 (4 votes)

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 shaving the sphere | Comment 15 of 21 |
(In reply to re(3): another solution by SilverKnight)

No, the correct solution is 13/3. This was best alluded to before visually by the intersection of the hyperboloid xy + xz + yz = 3 and the plane x + y + z = 5. It becomes much easier to see in examining the intersection of the plane x + y + z = 5 and the sphere x^2 + y^2 + z^2 = 19.

In this case we can clearly see that the solution set to both equations is the circle centered at 5/3,5/3,5/3 with radius √(19 - 25/3) = √32/3. Of course, this exists in the plane with the points 5,0,0 0,5,0 and 0,0,5. If we wish to maximize x we simply need to choose the point on the circle closest to 5,0,0. With some help from pythagorus we see that point is (13/3,1/3,1/3).
 Posted by Eric on 2003-11-18 21:24:48

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