Question Number 32629 by rahul 19 last updated on 29/Mar/18
$$\boldsymbol{{I}}{f}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{9}\:,\:\mathrm{4}{a}^{\mathrm{2}} +\mathrm{9}{b}^{\mathrm{2}} =\mathrm{16}, \\ $$$${then}\:\boldsymbol{{maximum}}\:{value}\:{of}\: \\ $$$$\mathrm{4}{a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{9}{b}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{12}{abxy}\:{is}\:? \\ $$
Answered by MJS last updated on 31/Mar/18
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{9}\:\Rightarrow\:\mid{x}\mid\leqslant\mathrm{3}\:\wedge\:\mid{y}\mid\leqslant\mathrm{3} \\ $$$$\mathrm{4}{a}^{\mathrm{2}} +\mathrm{9}{b}^{\mathrm{2}} =\mathrm{16}\:\Rightarrow\:\mid{a}\mid\leqslant\mathrm{2}\:\wedge\:\mid{b}\mid\leqslant\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\mathrm{4}{a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{9}{b}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{12}{abxy}=\left(\mathrm{2}{ax}−\mathrm{3}{by}\right)^{\mathrm{2}} \\ $$$${y}=\pm\sqrt{\mathrm{9}−{x}^{\mathrm{2}} } \\ $$$${b}=\pm\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\mathrm{4}−{a}^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}{ax}\pm\mathrm{2}\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }\right)^{\mathrm{2}} = \\ $$$$=\mathrm{4}\left({ax}\pm\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$${f}\left({x}\right)={ax}+\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }\sqrt{\mathrm{9}−{x}^{\mathrm{2}} } \\ $$$${f}'\left({x}\right)={a}−\frac{{x}\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}{\:\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }} \\ $$$${f}'\left({x}\right)=\mathrm{0}\:\Rightarrow\:{x}=\pm\frac{\mathrm{3}}{\mathrm{2}}{a} \\ $$$$\mathrm{4}\left({ax}\pm\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }\right)^{\mathrm{2}} = \\ $$$$=\mathrm{4}\left(\pm\frac{\mathrm{3}}{\mathrm{2}}{a}^{\mathrm{2}} \pm\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{4}−{a}^{\mathrm{2}} \right)\right)^{\mathrm{2}} = \\ $$$$=\mathrm{4}\left(\pm\mathrm{6}\right)^{\mathrm{2}} \:\vee\:\mathrm{4}\left(\pm\left(\mathrm{6}−\mathrm{3}{a}^{\mathrm{2}} \right)\right)^{\mathrm{2}} = \\ $$$$=\mathrm{144}\:\vee\:\mathrm{36}\left(\mathrm{2}−{a}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${g}\left({a}\right)=\mathrm{36}\left(\mathrm{2}−{a}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${g}'\left({a}\right)=\mathrm{144}{a}\left(\mathrm{2}−{a}^{\mathrm{2}} \right) \\ $$$${g}'\left({a}\right)=\mathrm{0}\:\Rightarrow\:{a}=\mathrm{0}\:\vee\:{a}=\pm\sqrt{\mathrm{2}} \\ $$$${g}\left(\pm\sqrt{\mathrm{2}}\right)=\mathrm{0} \\ $$$${g}\left(\mathrm{0}\right)=\mathrm{144} \\ $$$$\mathrm{Answer}=\mathrm{144} \\ $$$${a}=\mathrm{0} \\ $$$${b}=\pm\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${x}=\mathrm{0} \\ $$$${y}=\pm\mathrm{3} \\ $$
Commented by rahul 19 last updated on 31/Mar/18
$${thank}\:{u}\:{sir}\:! \\ $$