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Maximum Value (Posted on 2002-06-19) Difficulty: 3 of 5
We have :
      x^2+xy+y^2=3 and
      y^2+yz+z^2=16
      A=xy+yz+zx
Find the maximum value of A. Find x, y and z when A=max value.

(Remember the category)

See The Solution Submitted by vohonam    
Rating: 3.2857 (7 votes)

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Solution Solution: Part 3 Comment 18 of 18 |
(In reply to Solution: Part 2 by Brian Smith)

From Part 2:

217*y^4 - 608*y^2 + 256 = 0

Solving this equation yields y = 4/sqrt 31, y = -4/sqrt 31, y = 4/sqrt 7, and y = -4/sqrt 7 as possible values of y which yield a maximum value of A.

Since all the sign variables dropped out, each y has two possible x's and two possible z's

If y = 4/sqrt 31, then x = -11/sqrt 31 or 7/sqrt 31 and z = 20/sqrt 31 or -24/sqrt 31

If y = -4/sqrt 31, then x = 11/sqrt 31 or -7/sqrt 31 and z = -20/sqrt 31 or 24/sqrt 31

If y = 4/sqrt 7, then x = 1/sqrt 7 or -5/sqrt 7 and z = 8/sqrt 7 or -12/sqrt 7

If y = -4/sqrt 7, then x = -1/sqrt 7 or 5/sqrt 7 and z = -8/sqrt 7 or 12/sqrt 7

From Part 1: y = 2 and y = -2 are possible values for a maximum.

If y = 2, then x = -1 and z = -1+sqrt 13 or -1-sqrt 13

If y = -2, then x = 1 and z = 1+sqrt 13 or 1-sqrt 13

In all there are 20 candidate points to test. The two which create the highest value A=8 are (y=4/sqrt 31, x=7/sqrt 31, z=20/sqrt 31) and (y=-4/sqrt 31, x=-7/sqrt 31, z=-20/sqrt 31).
  Posted by Brian Smith on 2003-09-04 12:01:53

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