Question

If $a^2{x}^4+b^2{y}^4=c^6$ then the maximum value of xy$(a>0,b>0,c>0)$ is:

• A. $\frac{c^3}{2ab}$
• B. $\frac{c^3}{\sqrt{2ab}}$
• C. $\frac{c^3}{ab}$
• D. $\frac{c^3}{\sqrt{ab}}$

Answer 1

Answer: (B)

$\frac{c^3}{\sqrt{2ab}}$

$a^2{x}^4+b^2{y}^4=c^6,a>0,b>0,c>0A.M\ge G.M\frac{a^2{x}^4+b^2{y}^4}{2}\ge \sqrt{\left(a^2{x}^4\right)\left(b^2{y}^4\right)}ab{x}^2{y}^2\le \frac{c^6}{2}{x}^2{y}^2\le \frac{c^6}{2ab}xy\le \frac{c^3}{\sqrt{2ab}}(a>0,b>0,c>0)$

Minimum value of $xy=\frac{c^3}{\sqrt{2ab}}$