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Question Number 121353 by john santu last updated on 06/Nov/20

Answered by liberty last updated on 07/Nov/20

$$\mathrm{Let}\mathrm{x}\mathrm{be}\mathrm{the}\mathrm{x}-\mathrm{coordinate}\mathrm{of}\mathrm{end}\mathrm{point}$$$$\mathrm{that}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{x}-\mathrm{axis}\mathrm{and}\mathrm{let}\mathrm{the}\mathrm{other}$$$$\mathrm{endpoint}\mathrm{be}(0,\mathrm{y}).\mathrm{Thus}\mathrm{there}\mathrm{is}\mathrm{a}\mathrm{function}$$$$\mathrm{f}\mathrm{which}\mathrm{gives}\mathrm{y}\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{x}.\mathrm{Since}$$$${\mathrm{x}}^2+{\mathrm{y}}^2=1\mathrm{we}\mathrm{have}\mathrm{f}\left(\mathrm{x}\right)=\sqrt{1-{\mathrm{x}}^2}(0⩽\mathrm{x}⩽1)$$$$\mathrm{and}\mathrm{for}\mathrm{each}\mathrm{x},\mathrm{the}\mathrm{area}\mathrm{enclosed}\mathrm{is}$$$$\mathrm{A}\left(\mathrm{x}\right)=\frac{1}{2}xf\left(x\right)=\frac{1}{2}x\sqrt{1-x^2}$$$$weneedinvestigasithefunctionA$$$$\mathrm{for}\mathrm{maxima}.\mathrm{Now}{\mathrm{A}}^{\prime }\left(\mathrm{x}\right)=\frac{1}{2}[\frac{-{\mathrm{x}}^2}{\sqrt{1-{\mathrm{x}}^2}}+\sqrt{1-{\mathrm{x}}^2}]$$$${\mathrm{A}}^{\prime }\left(\mathrm{x}\right)=-\frac{1}{2}\frac{{2\mathrm{x}}^2-1}{\sqrt{1-{\mathrm{x}}^2}}=0$$$$\mathrm{when}x=\pm \frac{\sqrt{2}}{2}.\mathrm{Since}\mathrm{we}\mathrm{are}\mathrm{concerned}$$$$\mathrm{only}\mathrm{with}\mathrm{numbers}\mathrm{on}\mathrm{interval}[0,1]$$$$\mathrm{only}x=\frac{\sqrt{2}}{2}.H\mathrm{ere}\mathrm{A}=\frac{1}{4}\mathrm{is}\mathrm{maximum}$$$$\mathrm{area}\mathrm{of}\mathrm{triangle}$$

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