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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.

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

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Some Thoughts Most of a solution | Comment 12 of 18 |
I remember solving a simpler problem like this in highschool calculus.

Solve the first two equations leaving y as a parameter:
x = sqrt( 3 - 3(y^2)/4 ) - y/2
z = sqrt( 16 - 3(y^2)/4 ) - y/2

Substitute into A:
A = y*(sqrt (3 - 3*(y^2)/4) - y/2)
+ y*(sqrt (16 - 3*(y^2)/4) - y/2)
+ (sqrt (16 - 3*(y^2)/4) - y/2)*(sqrt (3 - 3*(y^2)/4) - y/2)

Now things get messy, find the first derivative of A. I had to use a symbolic calculator to make sure I got this right.

dA/dy = (-3*x/(4*sqrt(3-3*x^2/4))-1/2)*(sqrt(16-3*x^2/4)-x/2) +(sqrt(3-3*x^2/4)-x/2)*(-3*x/(4*sqrt(16-3*x^2/4))-1/2) +x*(-3*x/(4*sqrt(16-3*x^2/4))-1/2)+x*(-3*x/(4*sqrt(3-3*x^2/4))-1/2) +sqrt(16-3*x^2/4)+sqrt(3-3*x^2/4)-x

Numerically solving dA/dy = 0 yeilds A = 8.000000 and y = 0.718421 = sqrt(16/31) as the only maximum.
  Posted by Brian Smith on 2003-04-08 09:59:54
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