Question

If $a^2{x}^4+b^2{y}^4=c^6,$ then maximum value of $xy$ 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)

Given, $a^2{x}^4+b^2{y}^4=c^6$
$\Rightarrow$ $y^4=\frac{c^6-a^2{x}^4}{b^2}$
$\Rightarrow$ $y={\left(\frac{c^6-a^2{x}^4}{b^2}\right)}^{1/4}$
Let $z=xy=x{\left(\frac{c^6-a^2{x}^4}{b^2}\right)}^{1/4}$
$\Rightarrow$ $z={\left(\frac{c^6{x}^4-a^2{x}^6}{b^2}\right)}^{\frac{1}{4}}$
$\Rightarrow$ $\frac{dz}{dx}=\frac{1}{4}{\left(\frac{c^6{x}^4-a^2{x}^8}{b^2}\right)}^{-3/4}.\left(\frac{4x^3{c}^6}{b^2}-\frac{8a^2{x}^7}{b^2}\right)$
For maxima and minima, $\frac{dy}{dx}=0$
$\Rightarrow$ $\frac{1}{4}{\left(\frac{c^6{x}^4-a^2{x}^8}{b^2}\right)}^{-3/4}\times \left(\frac{4x^3{c}^6}{b^2}-\frac{8a^2{x}^7}{b^2}\right)=0$
$\Rightarrow$ $\frac{4x^3{c}^6}{b^2}-\frac{8a^2{x}^7}{b^2}=0$
$\Rightarrow$ $\frac{4x^3}{b^2}(c^6-2a^2{x}^4)=0$
$\Rightarrow$ $c^6-2a^2{x}^4=0$
$\Rightarrow$ $x^4=\frac{c^6}{2a^2}$
$\Rightarrow$ $x=\pm \frac{c^{6/4}}{2^{1/4}{a}^{2/4}}=\pm \frac{c^{3/2}}{2^{1/4}.a^{1/2}}$
$\therefore$ At $x=\pm \frac{c^{3/2}}{2^{1/4}\sqrt{a}},z$ is maximum
$\therefore$ $z={\left[\frac{c^6}{b^2}\left(\frac{c^6}{2a^2}\right)-\frac{a^2}{b^2}{\left(\frac{c^6}{2a^2}\right)}^2\right]}^{1/4}$
$={\left[\frac{c^{12}}{2a^2{b}^2}.\frac{c^{12}}{4a^2{b}^2}\right]}^{1/4}$
$={\left(\frac{c^{12}}{4a^2{b}^2}\right)}^{1/4}$ $=\frac{c^3}{\sqrt{2ab}}$