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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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(5830) (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.
==> ==> (because the function y = x^3 is 1-to-1.)
==> .
Hence the maximum value of xyz is 1/162, and it happens if and only if x = y = z = .
For the second problem,
v = 1- u, so that
==> ==> u = 0, 2/3. There is no need to test for u = 0.
when u = 2/3.
==> there is absolute max when u = 2/3, by the 2nd derivative test.
==> the max value of is .
