Question

# The greatest value of x2y3 where x>0, y>0 and 3x+4y=5 is 316386595

Solution

## The correct option is A 316Since, x>0, y>0 so using AM-GM inequality, we have 2⋅3x2+3⋅4y35≥((3x2)2(4y3)3)1/5 ⇒x2y3≤316

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