Find the greatest value of `x^2y^3, w h e r exa n dy` lie in the first quadrant on the line `3x+4y=5.`

Explanation

Given that 3x + 4y = 5
since we have expression `x^(2) y^(3)` we consider
`2((3x)/(2)) + 3 ((4y)/(3)) = 3x + 4y = 5`
Using A.M `ge` G.M for weighted means, we get
`(2((3x)/(2)) + 3 ((4)/(3)))/(2 + 3) ge [((3x)/(2))^(2) ((4y)/(3))^(3)]^((1)/(5))`
`implies ((3x + 4y)/(5))^(5) ge ((3)/(2))^(2) ((4)/(3))^(3) x^(2) y^(3)`
`implies x^(2) y^(2) le ((2)/(3))^(2) ((3)/(4))^(3)`
`implies x^(2) y^(3) le (3)/(16)`
so, the greatest value of `x^(2) y^(3)` is `(3)/(16)`