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Question

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

A

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

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B

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

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C

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

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D

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

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Solution

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}$$⇒$ $y={\left(\frac{{c}^{6}-{a}^{2}{x}^{4}}{{b}^{2}}\right)}^{1}{4}}$Now, $\begin{array}{rcl}f\left(x\right)& =& xy\\ & =& x{\left(\frac{{c}^{6}-{a}^{2}{x}^{4}}{{b}^{2}}\right)}^{1}{4}}\\ & =& {\left(\frac{{c}^{6}{x}^{4}-{a}^{2}{x}^{8}}{{b}^{2}}\right)}^{1}{4}}\end{array}$Step 2. Differentiate$f\left(x\right)$ with respect to $x$${f}^{\text{'}}\left(x\right)=\frac{1}{4}{\left(\frac{{c}^{6}{x}^{4}-{a}^{2}{x}^{8}}{{b}^{2}}\right)}^{-3}{4}}\left(\frac{4{x}^{3}{c}^{6}-8{a}^{2}{x}^{7}}{{b}^{2}}\right)\phantom{\rule{0ex}{0ex}}$Put ${f}^{\text{'}}\left(x\right)=0$$\left(\frac{4{x}^{3}{c}^{6}-8{a}^{2}{x}^{7}}{{b}^{2}}\right)=0$$⇒$ ${x}^{4}=\frac{{c}^{6}}{2{a}^{2}}$$⇒$ $x=±\frac{{c}^{3}{2}}}{{2}^{1}{4}}\sqrt{a}}$Step 3. At $x=\frac{{c}^{3}{2}}}{{2}^{1}{4}}\sqrt{a}}$, $f\left(x\right)$ is maximum and hence$\begin{array}{rcl}f\left(\frac{{c}^{3}{2}}}{{2}^{1}{2}}\sqrt{a}}\right)& =& {\left(\frac{{c}^{12}}{2{a}^{2}{b}^{2}}-\frac{{c}^{12}}{4{a}^{2}{b}^{2}}\right)}^{1}{4}}\\ & =& {\left(\frac{{c}^{12}}{4{a}^{2}{b}^{2}}\right)}^{1}{4}}\\ & =& \frac{{\mathbf{c}}^{\mathbf{3}}}{\sqrt{\mathbf{a}\mathbf{b}}}\end{array}$Hence, Option ‘C’ is Correct.Alternative method:We know thatArithmetic mean $\ge$ Geometric mean i.e.$AM\ge GM$$⇒$$\frac{{a}^{2}{x}^{4}+{b}^{2}{y}^{4}}{2}\ge {\left({a}^{2}{x}^{4}×{b}^{2}{y}^{4}\right)}^{\frac{1}{2}}$$⇒$ $\frac{{c}^{6}}{2}\ge a{x}^{2}b{y}^{2}$$⇒$ $\frac{{c}^{6}}{2}\ge ab{\left(xy\right)}^{2}$$⇒$ ${\left(xy\right)}^{2}\le \frac{{c}^{6}}{2ab}$Thus, $\frac{-{c}^{3}}{\sqrt{2ab}}\le xy\le \frac{{c}^{3}}{\sqrt{2ab}}$Hence, Option ‘C’ is Correct.

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