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

Options

A. `(c^(2))/(sqrt(ab)`
B. `(c^(3))/(sqrt(ab)`
C. `(c^(3))/(sqrt(2ab)`
D. `(c^(3))/(sqrt(2ab)`

Answer

3

Explanation

`a^(2)x^(4)+b^(2)y^(4)=c^(6)`
or `y=(c^(6)-a^(2)x^(4))/(b^(2))^(1//4)`
or `f(X)=xy=x(c^(6)-a^(2)x^(4))/(b^(2))^(1//4)`
`=(c^(6)x^(4)-a^(2)x^(8))/(b^(2))^(1//4)`
Differentiate f(X) w.r.t we get
`f(X)=1/4(c^(6)x^(4)-a^(2)x^(8))/(b^(2))^(3//4)(4x^(3)c^(6))/(b^(2))-(8x^(7)a^(2))/(b^(2))`
`f(X)=0 rarrr (4x^(3)c^(6))/(b^(2))-(8x^(7)a^(2))/(b^(2))=0`
`or x^(4)=(c^(6))/(2a^(2))or x=pm(c^(3//2))/(2^(1//4)sqrt(a))`
At x `=(c^(3//2))/(2^(1//4)sqrt(a))`, f(X) will be maximum so
`f(c^(3//2))/(2^(1//4sqrt(a))=(c^(1//4))/(2a^(2)b^(2))-(c^(12))/(4a^(2)b^(2)(1//4))=(c^(12))/(4a^(2)b^(2)(1//4))`