Find the greatest value of `x^3y^4` if `2x + 3y = 7` and `x>=0,y>=0` .

Explanation

To find the greatest value of `x^(3)y^(4)` or `(x)(x)(x)(y)(y)(y)(y)`
Hence, x reeats 3 times and y repeats 4 tims.
Given, `2x+3y=7`,
then multiplying and dividing coefficients of x and y by 3 and 4, respectively.
Rwrite `3((2x)/(3))+4((3y)/(4))=7`
or `((2x)/(3))+((2x)/(3))+((2x)/(3))+((3y)/(4))+((3y)/(4))+((3y)/(4))+((3y)/(4))=7`
Hence, `k=7` and `n=7`
Hence, greatest value of
`((2x)/(3))((2x)/(3))((2x)/(3))((3y)/(4))((3y)/(4))((3y)/(4))((3y)/(4))` is `((7)/(7))^(7)`.
or greatest value of `(2^(3)*3^(4))/(3^(3)*4^(4)) x^(3)y^(4)` is 1.
Thus, greatest value of `x^(3)y^(4)` is (32)/(3)`.