Question

If $a^2{x}^4+b^2{y}^4=c^6$,then 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 explanation for the correct option

Step 1. Find the maximum value of $xy$:

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

$\Rightarrow$ $y={\left(\frac{c^6-a^2{x}^4}{b^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$

Now,

$\begin{array}{rcl}f\left(x\right)& =& xy\\ & =& x{\left(\frac{c^6-a^2{x}^4}{b^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\\ & =& {\left(\frac{c^6{x}^4-a^2{x}^8}{b^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\end{array}$

Step 2. Differentiate$f\left(x\right)$ with respect to $x$

$f^{\text{'}}\left(x\right)=\frac{1}{4}{\left(\frac{c^6{x}^4-a^2{x}^8}{b^2}\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\left(\frac{4x^3{c}^6-8a^2{x}^7}{b^2}\right)$

Put $f^{\text{'}}\left(x\right)=0$

$\left(\frac{4x^3{c}^6-8a^2{x}^7}{b^2}\right)=0$

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

$\Rightarrow$ $x=\pm \frac{c^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{2^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}\sqrt{a}}$

Step 3. At $x=\frac{c^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{2^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}\sqrt{a}}$, $f\left(x\right)$ is maximum and hence

$\begin{array}{rcl}f\left(\frac{c^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{2^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\sqrt{a}}\right)& =& {\left(\frac{c^{12}}{2a^2{b}^2}-\frac{c^{12}}{4a^2{b}^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\\ & =& {\left(\frac{c^{12}}{4a^2{b}^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\\ & =& \frac{{\mathbf{c}}^{\mathbf{3}}}{\sqrt{\mathbf{a}\mathbf{b}}}\end{array}$

Hence, Option ‘C’ is Correct.

Alternative method:

We know that

Arithmetic mean $\ge$ Geometric mean i.e.$AM\ge GM$

$\Rightarrow$$\frac{a^2{x}^4+b^2{y}^4}{2}\ge {\left(a^2{x}^4\times b^2{y}^4\right)}^{\frac{1}{2}}$

$\Rightarrow$ $\frac{c^6}{2}\ge a{x}^{2}b{y}^2$

$\Rightarrow$ $\frac{c^6}{2}\ge ab{\left(xy\right)}^2$

$\Rightarrow$ ${\left(xy\right)}^2\le \frac{c^6}{2ab}$

Thus, $\frac{-c^3}{\sqrt{2ab}}\le xy\le \frac{c^3}{\sqrt{2ab}}$

Hence, Option ‘C’ is Correct.