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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 2 | Comment 17 of 18 |
(In reply to Solution: Part 1 by Brian Smith)

From Part 1:

(1/4)*( -6*y + s*sqrt[12 - 3*y^2] + t*sqrt[64 - 3*y^2] + y*(-6*s*y)/(2*sqrt[12 - 3*y^2]) + y*(-6*t*y)/(2*sqrt[64 - 3*y^2]) + s*sqrt[12 - 3*y^2]*(-6*t*y)/(2*sqrt[64 - 3*y^2]) + t*sqrt[64 - 3*y^2]*(-6*s*y)/(2*sqrt[12 - 3*y^2]) ) = 0

Now to solve this messy equation.
First multiply each side by 4*s*sqrt[12 - 3*y^2]*t*sqrt[64 - 3*y^2]
Note: s*s = 1 and t*t = 1

0 = -6*y*s*t*sqrt[12 - 3*y^2]*sqrt[64 - 3*y^2] + (12 - 3*y^2)*t*sqrt[64 - 3*y^2] + (64 - 3*y^2)*s*sqrt[12 - 3*y^2] + -3*y^2*s*sqrt[12 - 3*y^2] + -3*y^2*t*sqrt[64 - 3*y^2] + -3*y*(12 - 3*y^2) + -3*y*(64 - 3*y^2)

After some algebra and rearranging some terms:

6*y*s*t*sqrt[12 - 3*y^2]*sqrt[64 - 3*y^2] + 228*y + -18*y^3 = (12 - 6*y^2)*t*sqrt[64 - 3*y^2] + (64 - 6*y^2)*s*sqrt[12 - 3*y^2]

After squaring each side:

(228*y - 18*y^3)^2 + 2*(228*y - 18*y^3)*(6*y*s*t*sqrt[12 - 3*y^2]*sqrt[64 - 3*y^2]) + 36*y^2*(12 - 3*y^2)*(64 - 3*y^2) = (12 - 6*y^2)^2*(64 - 3*y^2) + (64 - 6*y^2)^2*(12 - 3*y^2) + 2*(12 - 6*y^2)*(64 - 6*y^2)*s*t*sqrt[12 - 3*y^2]*sqrt[64 - 3*y^2]

After even more algebra and removing a common multiple: (It seems most calculus is 80% algebra)

9*y^6 - 228*y^4 + 1154*y^2 - 608 = (3*y^4 - 38*y^2 + 16)*s*t*sqrt[12 - 3*y^2]*sqrt[64 - 3*y^2]

After squaring again and still more algebra, a final equation is formed (the higher exponents dropped out):

217*y^4 - 608*y^2 + 256 = 0
  Posted by Brian Smith on 2003-09-04 11:46:19

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