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Question Number 32629 by rahul 19 last updated on 29/Mar/18

$$\mathit{I}f{x}^2+y^2=9,4a^2+9b^2=16,$$$$then\mathit{m}\mathit{a}\mathit{x}\mathit{i}\mathit{m}\mathit{u}\mathit{m}valueof$$$$4a^2{x}^2+9b^2{y}^2-12abxyis?$$

Answered by MJS last updated on 31/Mar/18

$$x^2+y^2=9\Rightarrow \mid x\mid ⩽3\wedge \mid y\mid ⩽3$$$$4a^2+9b^2=16\Rightarrow \mid a\mid ⩽2\wedge \mid b\mid ⩽\frac{4}{3}$$$$4a^2{x}^2+9b^2{y}^2-12abxy={(2ax-3by)}^2$$$$y=\pm \sqrt{9-x^2}$$$$b=\pm \frac{2}{3}\sqrt{4-a^2}$$$${(2ax\pm 2\sqrt{4-a^2}\sqrt{9-x^2})}^2=$$$$=4{(ax\pm \sqrt{4-a^2}\sqrt{9-x^2})}^2$$$$f\left(x\right)=ax+\sqrt{4-a^2}\sqrt{9-x^2}$$$$f^{\prime }\left(x\right)=a-\frac{x\sqrt{4-a^2}}{\sqrt{9-x^2}}$$$$f^{\prime }\left(x\right)=0\Rightarrow x=\pm \frac{3}{2}a$$$$4{(ax\pm \sqrt{4-a^2}\sqrt{9-x^2})}^2=$$$$=4{(\pm \frac{3}{2}{a}^2\pm \frac{3}{2}(4-a^2))}^2=$$$$=4{(\pm 6)}^2\vee 4{(\pm (6-3a^2))}^2=$$$$=144\vee 36{(2-a^2)}^2$$$$g\left(a\right)=36{(2-a^2)}^2$$$$g^{\prime }\left(a\right)=144a(2-a^2)$$$$g^{\prime }\left(a\right)=0\Rightarrow a=0\vee a=\pm \sqrt{2}$$$$g(\pm \sqrt{2})=0$$$$g\left(0\right)=144$$$$\mathrm{Answer}=144$$$$a=0$$$$b=\pm \frac{4}{3}$$$$x=0$$$$y=\pm 3$$

Commented by rahul 19 last updated on 31/Mar/18

$$thankusir!$$

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