Question

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

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

$a^2{x}^4+b^2{y}^4=c^6$
$\Rightarrow y^4=\frac{c^6-a^2{x}^4}{b^2}$
Let $s=xy$
$\Rightarrow s^4=x^4\left(\frac{c^6-a^2{x}^4}{b^2}\right)$
$\Rightarrow 4s^3\frac{ds}{dx}=\frac{{4c}^6{x}^3-{8a}^2{x}^7}{b^2}$
For maxima or minima,
$\frac{ds}{dx}=0$
$\Rightarrow 4x^3(c^6-2a^2{x}^4)=0$
$\Rightarrow x=0,x=\frac{c^{\frac{3}{2}}}{2^{\frac{1}{4}}{a}^{\frac{1}{2}}}$
$\frac{d^{2}s}{d{x}^2}<0atx=\frac{c^{\frac{3}{2}}}{2^{\frac{1}{4}}{a}^{\frac{1}{2}}}$
Hence there is a maximum at $x=\frac{c^{\frac{3}{2}}}{2^{\frac{1}{4}}{a}^{\frac{1}{2}}}$
Maximum value $=xy={\left(\frac{c^{12}}{{4a}^2{b}^2}\right)}^{\frac{1}{4}}=\frac{c^3}{\sqrt{2ab}}$