SOLUTION: hi am stuck on a question on theory of numbers If x+2y+3z = 1, where x, y, z are positive numbers. find the maximum value of xyz. If u+v=1, where u, v are positive numbers,find

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Question 461623: hi am stuck on a question on theory of numbers
If x+2y+3z = 1, where x, y, z are positive numbers. find the maximum value of xyz.
If u+v=1, where u, v are positive numbers,find the maximum value of uČv.
Answer by robertb(4012) About Me  (Show Source):
You can put this solution on YOUR website!
The first problem can be solved by the AM-GM inequality. I will solve the second problem using calculus.
1%2F3+=+%28x%2B2y+%2B+3z%29%2F3+%3E=+root%283%2C+x%2A2y%2A3z%29+=+root%283%2C+6xyz%29
==> 1%2F3+%3E=+root%283%2C+6xyz%29 ==> 1%2F27+%3E=+6xyz (because the function y = x^3 is 1-to-1.)
==> 1%2F162+%3E=+xyz.
Hence the maximum value of xyz is 1/162, and it happens if and only if x = y = z = %281%2F3%29%2Aroot%283%2C+1%2F6%29.
For the second problem,
v = 1- u, so that u%5E2v+=+u%5E2%281+-+u%29+=+u%5E2+-+u%5E3
==> d%28u%5E2+-+u%5E3%29%2Fdu+=+2u+-+3u%5E2+=+0 ==> u = 0, 2/3. There is no need to test for u = 0.
d%5E2%28u%5E2+-+u%5E3%29%2Fdu%5E2+=+2+-+6u+=+2+-+6%2A%282%2F3%29+=+-2+%3C0 when u = 2/3.
==> there is absolute max when u = 2/3, by the 2nd derivative test.
==> the max value of u%5E2v is %282%2F3%29%5E2%2A%281%2F3%29+=+4%2F27.