If `a^2x^4+b^2y^4=c^6`, then the maximum value of `xy` is (a) `c^3/(2ab)` (b) `c^3/sqrt(2ab)` (c) `c^3/(ab)` (d) `c^3/sqrt(ab)`

Explanation

Here, `a^2x^4 ge 0` and `b^2y^4 ge 0`

Arithmatic mean of `a^2x^4` and `b^2y^4 = 1/2(a^2x^4+b^2y^4)`

Geometric mean of `a^2x^4` and `b^2y^4 = (a^2x^4b^2y^4)^(1/2)`

We know, arithmatic mean of two numbers is always greater than or equal to their geometric mean.

`:. 1/2(a^2x^4+b^2y^4) ge (a^2x^4b^2y^4)^(1/2)`

`=>c^6/2 ge (a^2x^4b^2y^4)^(1/2)`

`=> c^12/4 ge (a^2x^4b^2y^4)`

`=>x^4y^4 le c^12/(4a^2b^2)`

...