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)
(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
Solution: Part 2
