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Question Number 155203 by nadovic last updated on 26/Sep/21

Answered by TheHoneyCat last updated on 29/Sep/21

let O be the center of the circle  let A be a vertice of the rectangle  let θ be an angle (in radian) between one of  the two axis (Oz or Oy) and the (OA) line    The dimentions of the rectangle are  (2cosθ,2sinθ) (or the opposite depending  on your definition)    Hence the area A of the rectangle is :  A=4∣cosθ.sinθ∣  let a=cos×sin  ∣a∣ is a π/2 periodical function  also on [0,π/2] ∣a∣=a so let us sudy the  restriction of a to this set:    (da/dθ)=((d cos)/dθ)sin + ((d sin)/dθ)cos  =−sin^2 +cos^2   so on this restriction  a′(θ)=0 ⇔ (cosθ)^2 =(sinθ)^2   ⇔cosθ=sinθ (caus′ they are positiv on the   interval)  ⇔θ=^π /_4   ⇔cosθ=sinθ=(√2)/2  ⇔a(θ)=(1/2)      Hence, the maximum area of the rectangle  is that of a square ie:  A=2

$$\mathrm{let}\:\mathrm{O}\:\mathrm{be}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle} \\ $$$$\mathrm{let}\:\mathrm{A}\:\mathrm{be}\:\mathrm{a}\:\mathrm{vertice}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rectangle} \\ $$$$\mathrm{let}\:\theta\:\mathrm{be}\:\mathrm{an}\:\mathrm{angle}\:\left(\mathrm{in}\:\mathrm{radian}\right)\:\mathrm{between}\:\mathrm{one}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{axis}\:\left(\mathrm{O}{z}\:\mathrm{or}\:\mathrm{O}{y}\right)\:\mathrm{and}\:\mathrm{the}\:\left(\mathrm{OA}\right)\:\mathrm{line} \\ $$$$ \\ $$$$\mathrm{The}\:\mathrm{dimentions}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rectangle}\:\mathrm{are} \\ $$$$\left(\mathrm{2cos}\theta,\mathrm{2sin}\theta\right)\:\left({or}\:{the}\:{opposite}\:{depending}\right. \\ $$$$\left.{on}\:{your}\:{definition}\right) \\ $$$$ \\ $$$$\mathrm{Hence}\:\mathrm{the}\:\mathrm{area}\:\mathscr{A}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rectangle}\:\mathrm{is}\:: \\ $$$$\mathscr{A}=\mathrm{4}\mid\mathrm{cos}\theta.\mathrm{sin}\theta\mid \\ $$$$\mathrm{let}\:{a}=\mathrm{cos}×\mathrm{sin} \\ $$$$\mid{a}\mid\:\mathrm{is}\:\mathrm{a}\:\pi/\mathrm{2}\:\mathrm{periodical}\:\mathrm{function} \\ $$$$\mathrm{also}\:\mathrm{on}\:\left[\mathrm{0},\pi/\mathrm{2}\right]\:\mid{a}\mid={a}\:\mathrm{so}\:\mathrm{let}\:\mathrm{us}\:\mathrm{sudy}\:\mathrm{the} \\ $$$$\mathrm{restriction}\:\mathrm{of}\:\mathrm{a}\:\mathrm{to}\:\mathrm{this}\:\mathrm{set}: \\ $$$$ \\ $$$$\frac{\mathrm{d}{a}}{\mathrm{d}\theta}=\frac{\mathrm{d}\:\mathrm{cos}}{\mathrm{d}\theta}\mathrm{sin}\:+\:\frac{\mathrm{d}\:\mathrm{sin}}{\mathrm{d}\theta}\mathrm{cos} \\ $$$$=−\mathrm{sin}^{\mathrm{2}} +\mathrm{cos}^{\mathrm{2}} \\ $$$$\mathrm{so}\:\mathrm{on}\:\mathrm{this}\:\mathrm{restriction} \\ $$$${a}'\left(\theta\right)=\mathrm{0}\:\Leftrightarrow\:\left(\mathrm{cos}\theta\right)^{\mathrm{2}} =\left(\mathrm{sin}\theta\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{cos}\theta=\mathrm{sin}\theta\:\left({caus}'\:{they}\:{are}\:{positiv}\:{on}\:{the}\:\right. \\ $$$$\left.{interval}\right) \\ $$$$\Leftrightarrow\theta=^{\pi} /_{\mathrm{4}} \\ $$$$\Leftrightarrow\mathrm{cos}\theta=\mathrm{sin}\theta=\sqrt{\mathrm{2}}/\mathrm{2} \\ $$$$\Leftrightarrow{a}\left(\theta\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Hence},\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rectangle} \\ $$$$\mathrm{is}\:\mathrm{that}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}\:{ie}: \\ $$$$\mathscr{A}=\mathrm{2} \\ $$

Commented by nadovic last updated on 29/Sep/21

Sir, the area is given as 32 sq. in.

$${Sir},\:{the}\:{area}\:{is}\:{given}\:{as}\:\mathrm{32}\:{sq}.\:{in}. \\ $$

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