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Question

The greatest value of x2y3 where x>0, y>0 and 3x+4y=5 is 
  1. 316
  2. 38
  3. 65
  4. 95


Solution

The correct option is A 316
Since, x>0, y>0 so using AM-GM inequality, we have
23x2+34y35((3x2)2(4y3)3)1/5
x2y3316

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