document.write( "Question 461623: hi am stuck on a question on theory of numbers\r
\n" ); document.write( "\n" ); document.write( "If x+2y+3z = 1, where x, y, z are positive numbers. find the maximum value of xyz.
\n" ); document.write( "If u+v=1, where u, v are positive numbers,find the maximum value of u²v. \n" ); document.write( "

Algebra.Com's Answer #316670 by robertb(5830) $\"\"$ $\"About$

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The first problem can be solved by the AM-GM inequality. I will solve the second problem using calculus.\r
\n" ); document.write( "\n" ); document.write( " $\"1%2F3+=+%28x%2B2y+%2B+3z%29%2F3+%3E=+root%283%2C+x%2A2y%2A3z%29+=+root%283%2C+6xyz%29\"$ \r
\n" ); document.write( "\n" ); document.write( "==> $\"1%2F3+%3E=+root%283%2C+6xyz%29\"$ ==> $\"1%2F27+%3E=+6xyz\"$ (because the function y = x^3 is 1-to-1.)\r
\n" ); document.write( "\n" ); document.write( "==> $\"1%2F162+%3E=+xyz\"$ .\r
\n" ); document.write( "\n" ); document.write( "Hence the maximum value of xyz is 1/162, and it happens if and only if x = y = z = $\"%281%2F3%29%2Aroot%283%2C+1%2F6%29\"$ .\r
\n" ); document.write( "\n" ); document.write( "For the second problem, \r
\n" ); document.write( "\n" ); document.write( "v = 1- u, so that $\"u%5E2v+=+u%5E2%281+-+u%29+=+u%5E2+-+u%5E3\"$ \r
\n" ); document.write( "\n" ); document.write( "==> $\"d%28u%5E2+-+u%5E3%29%2Fdu+=+2u+-+3u%5E2+=+0\"$ ==> u = 0, 2/3. There is no need to test for u = 0.\r
\n" ); document.write( "\n" ); document.write( " $\"d%5E2%28u%5E2+-+u%5E3%29%2Fdu%5E2+=+2+-+6u+=+2+-+6%2A%282%2F3%29+=+-2+%3C0\"$ when u = 2/3.
\n" ); document.write( "==> there is absolute max when u = 2/3, by the 2nd derivative test.\r
\n" ); document.write( "\n" ); document.write( "==> the max value of $\"u%5E2v\"$ is $\"%282%2F3%29%5E2%2A%281%2F3%29+=+4%2F27\"$ . \n" ); document.write( "

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