Question

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

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

Answer 1

The correct option is C $\frac{c^3}{\sqrt{2ab}}$

The given equation is:

$a^2{x}^4+b^2{y}^4=c^6$

$\Rightarrow y^4=\frac{c^6-a^2{x}^4}{b^2}$

Let $z=xy$. To find the extremum points we differentiate and equate it to zero

$\Rightarrow z=x\left(\frac{c^6-a^2{x}^4}{b^2}\right)$

$\Rightarrow z^{\prime }=(\frac{c^6-a^2{x}^4}{b^2}{)}^{1/4}-\frac{x}{4}(\frac{c^6-a^2{x}^4}{b^2}{)}^{-3/4}.4x^3\frac{a^2}{b^2}$

$\Rightarrow z^{\prime }=(\frac{c^6-a^2{x}^4}{b^2}{)}^{1/4}-(\frac{c^6-a^2{x}^4}{b^2}{)}^{-3/4}.x^4\frac{a^2}{b^2}$

$z^{\prime }=0$

$\Rightarrow (\frac{c^6-a^2{x}^4}{b^2}{)}^{1/4}-(\frac{c^6-a^2{x}^4}{b^2}{)}^{-3/4}.x^4\frac{a^2}{b^2}=0$

$\Rightarrow (\frac{c^6-a^2{x}^4}{b^2}{)}^{1/4}-(\frac{c^6=a^2{x}^4}{b^2}{)}^{-3/4}.x^4\frac{a^2}{b^2}$

$\Rightarrow x^4=\frac{b^2}{a^2}\left(\frac{c^6-a^2{x}^4}{b^2}\right)$

$\Rightarrow x^4=\left(\frac{c^6-a^2{x}^4}{a^2}\right)$

$\Rightarrow x^4=\frac{c^6}{a^2}-x^4$

$\Rightarrow x^4=\frac{c^6}{2a^2}$

$\Rightarrow x=\frac{c^{6/4}}{2^{1/4}{a}^{1/2}}$

Substituting the x value in the above equation we get,

$\Rightarrow z=\frac{c^{6/4}}{2^{1/4}{a}^{1/2}}(\frac{c^6}{b^2}-\frac{a^2{c}^6}{2a^2{b}^2}{)}^{1/4}$

$\Rightarrow z=\frac{c^{6/4}}{2^{1/4}{a}^{1/2}}(\frac{c^6}{b^2}-\frac{c^6}{2b^2}{)}^{1/4}$

$\Rightarrow z=\frac{c^{6/4}}{2^{1/4}{a}^{1/2}}(\frac{c^6}{2b^2}{)}^{1/4}$

$\Rightarrow z=(\frac{c^{12}}{4a^2{b}^2}{)}^{1/4}$

$\Rightarrow z=\frac{c^3}{\sqrt{2ab}}$ .....Answer