# 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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 Click here to see ALL problems on Evaluation Word Problems 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)   (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 .