Find the least value of `3x+4y` for positive values of x and y, dubject to the condition `x^(2)y^(3)=6`.

Explanation

Given, `x^(2)y^(3)=6`
or `(x)(x)(y)(y)(y)=6`
Hence, x repeats 2 times and y repeats 3 tims.
`:. 3x+4y=2((3x)/(2))+3((4uy)/(3))`
`=((3x)/(2))+((3x)/(2))+((4y)/(3))+((4y)/(3))+((4y)/(3))`
multiplying and dividing coefficients of x and y by 2 and 3, respectively and write `x^(2)y^(3)=6`
`implies ((3x)/(2))+((3x)/(2))+((4y)/(3))+((4y)/(3))+((4y)/(3))=(3^(2))/(2^(2))xx(4^(3))/(3^(3))xx6=32`
Hence, `n=5` and `k=32`
Hence,least value of `(3x)/(2)+(3x)/(2)+(4y)/(3)+(4y)/(3)+(4y)/(3)`
`=5(32)^((1)/(5))=10`
i.e.least value of `3x+4y=10`.