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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 1 | Comment 16 of 18 |
First, express x and z in terms of y.

Let s and t be sign variables (equal to 1 or -1), since square roots can be positive or negative.

x = (-y/2) + s*sqrt[3 - (3*y^2/4)]
z = (-y/2) + t*sqrt[16 - (3*y^2/4)]

Now A can be expressed in terms of y and some sign variables.

A = ((-y/2) + s*sqrt[3 - (3*y^2/4)])*((-y/2) + t*sqrt[16 - (3*y^2/4)]) + y*((-y/2) + s*sqrt[3 - (3*y^2/4)]) + y*((-y/2) + t*sqrt[16 - (3*y^2/4)])

After some algebra:

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

The function A is defined as long as the square roots are real which means 12 - 3*y^2 >= 0 and 64 - 3*y^2 >= 0. From this, A is defined for 2 >= y >= -2.

The maximum value of A occurs at y = 2, y = -2, or some local maximum. Now its time for some calculus.

dA/dy = (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
  Posted by Brian Smith on 2003-09-04 11:15:27
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