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Question Number 128751 by bemath last updated on 10/Jan/21

$$\mathrm{Find}\mathrm{maximum}\mathrm{value}\mathrm{of}$$$$\mid \mathrm{x}\mid \sqrt{16-{\mathrm{y}}^2}+\mid \mathrm{y}\mid \sqrt{4-{\mathrm{x}}^2}\mathrm{for}\mathrm{all}\mathrm{real}\mathrm{x}\mathrm{and}\mathrm{y}.$$$$$$

Answered by mr W last updated on 10/Jan/21

$$a=\mid x\mid =2\sin \alpha ⩾0,0⩽\alpha ⩽\frac{\pi }{2}$$$$b=\mid y\mid =4\sin \beta ⩾0,0⩽\beta ⩽\frac{\pi }{2}$$$$a\sqrt{4^2-b^2}+b\sqrt{2^2-a^2}$$$$=8\sin \alpha \cos \beta +8\sin \beta \cos \alpha$$$$=8\sin (\alpha +\beta )$$$$max.=8at\sin (\alpha +\beta )=1,e.g.\alpha =\beta =\frac{\pi }{4}$$

Commented by liberty last updated on 10/Jan/21

$$\mathrm{i}\mathrm{think}\mathrm{not}8\mathrm{is}\mathrm{maximum}\mathrm{value}$$

Commented by mr W last updated on 10/Jan/21

$$ifthemaximumfrom8\sin (\alpha +\beta )$$$$isnot8,theni{don}^{\prime }tunterstandthis$$$$worldanymore:)$$

Commented by liberty last updated on 10/Jan/21

$$\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{reason}\mathrm{by}\mathrm{assuming}\mid \mathrm{x}\mid =2\sin \alpha ?$$

Commented by mr W last updated on 10/Jan/21

$$since4-x^2⩾0\Rightarrow \mid x\mid ⩽2$$

Commented by liberty last updated on 10/Jan/21

$$\mathrm{ok}.\mathrm{you}\mathrm{are}\mathrm{right}.$$

Answered by liberty last updated on 10/Jan/21

$$\mathrm{Using}\mathrm{the}\mathrm{inequality}xy⩽\frac{x^2+y^2}{2}$$$$\mathrm{we}\mathrm{have}\{\begin{array}{l}\mid \mathrm{x}\mid \sqrt{16-{\mathrm{y}}^2}⩽\frac{{\mathrm{x}}^2+16-{\mathrm{y}}^2}{2}\\ \mid \mathrm{y}\mid \sqrt{4-{\mathrm{x}}^2}⩽\frac{{\mathrm{y}}^2+4-{\mathrm{x}}^2}{2}\end{array}$$$$\mathrm{adding}\mathrm{the}\mathrm{both}\mathrm{equation}\mathrm{we}\mathrm{find}$$$$\mid \mathrm{x}\mid \sqrt{16-{\mathrm{y}}^2}+\mid \mathrm{y}\mid \sqrt{4-{\mathrm{x}}^2}⩽\frac{{\mathrm{x}}^2+16-{\mathrm{y}}^2+{\mathrm{y}}^2+4-{\mathrm{x}}^2}{2}=10$$$$\therefore \mathrm{its}\mathrm{maximum}\mathrm{value}\mathrm{is}10$$$$$$

Commented by mr W last updated on 10/Jan/21

$$canyousaywhatvaluesxandyhave$$$$suchthatyoureachthemaximum$$$$10?$$

Commented by liberty last updated on 10/Jan/21

$$10\mathrm{is}\mathrm{only}\mathrm{an}\mathrm{upper}\mathrm{bound}\mathrm{but}\mathrm{cannot}\mathrm{be}$$$$\mathrm{attained}\mathrm{since}\mathrm{the}\mathrm{inequality}\mathrm{xy}⩽\frac{{\mathrm{x}}^2+{\mathrm{y}}^2}{2}$$$$\mathrm{holds}\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}\mathrm{x}=\mathrm{y},\mathrm{therefore}\mathrm{the}\max$$$$\mathrm{value}\mathrm{is}\mathrm{attained}\mid \mathrm{x}\mid =\sqrt{16-{\mathrm{y}}^2}\mathrm{and}\mid \mathrm{y}\mid =\sqrt{4-{\mathrm{x}}^2}$$$$\mathrm{or}{\mathrm{x}}^2=16-{\mathrm{y}}^2\mathrm{and}{\mathrm{y}}^2=4-{\mathrm{x}}^2\iff {\mathrm{x}}^2+{\mathrm{y}}^2=16=4$$$$\mathrm{contradictory}!$$

Commented by mr W last updated on 10/Jan/21

$$correct!$$$$sobecarefulwhenaddingtwoor$$$$moreinequalitiesiftheyarenot$$$$independentfromeachother!$$

Answered by mr W last updated on 10/Jan/21

$$anotherway:$$$$f(a,b)=a\sqrt{16-b^2}+b\sqrt{4-a^2}$$$$\frac{\partial f}{\partial a}=\sqrt{16-b^2}-\frac{ab}{\sqrt{4-a^2}}=0$$$$\frac{\partial f}{\partial b}=-\frac{ab}{\sqrt{16-b^2}}+\sqrt{4-a^2}=0$$$$\Rightarrow ab=\sqrt{(4-a^2)(16-b^2)}$$$$\Rightarrow 4a^2+b^2=16$$$$\Rightarrow 16-b^2=4a^2$$$$\Rightarrow b=2\sqrt{4-a^2}$$$$\Rightarrow f_{max}=2a^2+2\sqrt{4-a^2}\times \sqrt{4-a^2}=8$$

Commented by liberty last updated on 10/Jan/21

$$\mathrm{yes}.\mathrm{nice}$$

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