Question Number 121353 by john santu last updated on 06/Nov/20
Answered by liberty last updated on 07/Nov/20
![Let x be the x−coordinate of end point that lies on the x−axis and let the other endpoint be (0,y). Thus there is a function f which gives y in terms of x. Since x^2 +y^2 =1 we have f(x)=(√(1−x^2 )) (0≤x≤1) and for each x , the area enclosed is A(x)=(1/2)xf(x) = (1/2)x(√(1−x^2 )) we need investigasi the function A for maxima.Now A′(x)=(1/2) [ ((−x^2 )/( (√(1−x^2 ))))+(√(1−x^2 )) ] A′(x)= −(1/2) ((2x^2 −1)/( (√(1−x^2 )))) = 0 when x = ± ((√2)/2) . Since we are concerned only with numbers on interval [ 0,1 ] only x = ((√2)/2). Here A = (1/4) is maximum area of triangle](https://www.tinkutara.com/question/Q121358.png)
$$\mathrm{Let}\:\mathrm{x}\:\mathrm{be}\:\mathrm{the}\:\mathrm{x}−\mathrm{coordinate}\:\mathrm{of}\:\mathrm{end}\:\mathrm{point}\: \\ $$$$\mathrm{that}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{x}−\mathrm{axis}\:\mathrm{and}\:\mathrm{let}\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{endpoint}\:\mathrm{be}\:\left(\mathrm{0},\mathrm{y}\right).\:\mathrm{Thus}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{function} \\ $$$$\mathrm{f}\:\mathrm{which}\:\mathrm{gives}\:\mathrm{y}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{x}.\:\mathrm{Since} \\ $$$$\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{1}\:\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:\:\left(\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{1}\right) \\ $$$$\mathrm{and}\:\mathrm{for}\:\mathrm{each}\:\mathrm{x}\:,\:\mathrm{the}\:\mathrm{area}\:\mathrm{enclosed}\:\mathrm{is}\: \\ $$$$\mathrm{A}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{xf}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${we}\:{need}\:{investigasi}\:{the}\:{function}\:{A} \\ $$$$\mathrm{for}\:\mathrm{maxima}.\mathrm{Now}\:\mathrm{A}'\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\left[\:\frac{−\mathrm{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}+\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:\right] \\ $$$$\mathrm{A}'\left(\mathrm{x}\right)=\:−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{2x}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\:\:=\:\mathrm{0} \\ $$$$\mathrm{when}\:{x}\:=\:\pm\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:.\:\mathrm{Since}\:\mathrm{we}\:\mathrm{are}\:\mathrm{concerned} \\ $$$$\mathrm{only}\:\mathrm{with}\:\mathrm{numbers}\:\mathrm{on}\:\mathrm{interval}\:\left[\:\mathrm{0},\mathrm{1}\:\right] \\ $$$$\mathrm{only}\:{x}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}.\:{H}\mathrm{ere}\:\mathrm{A}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{is}\:\mathrm{maximum} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{triangle} \\ $$