Question Number 28574 by ajfour last updated on 27/Jan/18
Commented by ajfour last updated on 27/Jan/18
$${Find}\:{entire}\:{green}\:{area}\:{in}\:{terms} \\ $$$${of}\:\boldsymbol{{r}},\:\boldsymbol{\theta}\:.\:{The}\:{two}\:{frames}\:{are} \\ $$$${square}\:{frames}\:{with}\:{common} \\ $$$${centre}.\:{Find}\:{also}\:{the}\:{maximum} \\ $$$${of}\:{the}\:{area}\:{for}\:{constant}\:\boldsymbol{{r}}. \\ $$
Commented by ajfour last updated on 29/Jan/18
$${mrW}\mathrm{2}\:{Sir},\:{would}\:{you}\:{solve}\:{this}? \\ $$
Answered by mrW2 last updated on 29/Jan/18
Commented by ajfour last updated on 29/Jan/18
$${thank}\:{you}\:{enormously}\:{Sir}! \\ $$$${extremely}\:{satisfactory}! \\ $$
Commented by mrW2 last updated on 29/Jan/18
![Eqn. of AB: x=(a/2) with a=(√2)r Eqn. of CD (=AB rotated by θ) x cos θ+y sin θ=(a/2) (a/2) cos θ+y_E sin θ=(a/2) ⇒y_E =((a(1−cos θ))/(2 sin θ)) x_F cos θ+(a/2) sin θ=(a/2) ⇒x_F =((a(1−sin θ))/(2 cos θ)) A_(ΔBEF) =(1/2)[(a/2)−((a(1−sin θ))/(2 cos θ))][(a/2)−((a(1−cos θ))/(2 sin θ))] A_(ΔBEF) =(a^2 /8)[((cos θ−1+sin θ)/(cos θ))][((sin θ−1+cos θ)/( sin θ))] A_(ΔBEF) =(a^2 /8)×(((cos θ+sin θ−1)^2 )/(sin θ cos θ)) A_(ΔBEF) =(a^2 /4)[1−((cos θ+sin θ−1)/(sin θ cos θ))] A_(Green) =4A_(ΔBEF) A_(Green) (θ)=a^2 [1−((cos θ+sin θ−1)/(sin θ cos θ))] A_(Green) (θ)=2r^2 [1−((cos θ+sin θ−1)/(sin θ cos θ))] with 0≤θ≤(π/2) A_(Green) ((π/2)+θ)=A_(Green) (θ) f(θ)=1−((cos θ+sin θ−1)/(sin θ cos θ)) f(θ)=1+((2[1−(√2) sin (θ+(π/4))])/(sin 2θ)) max. f(θ)=1+((2[1−(√2)])/1)=3−2(√2)≈0.17 at θ=(π/4) ⇒max.A_(Green) =(3−2(√2))a^2 ≈0.17a^2 ≈0.34r^2](https://www.tinkutara.com/question/Q28742.png)
$${Eqn}.\:{of}\:{AB}: \\ $$$${x}=\frac{{a}}{\mathrm{2}}\:{with}\:{a}=\sqrt{\mathrm{2}}{r} \\ $$$${Eqn}.\:{of}\:{CD}\:\left(={AB}\:{rotated}\:{by}\:\theta\right) \\ $$$${x}\:\mathrm{cos}\:\theta+{y}\:\mathrm{sin}\:\theta=\frac{{a}}{\mathrm{2}} \\ $$$$ \\ $$$$\frac{{a}}{\mathrm{2}}\:\mathrm{cos}\:\theta+{y}_{{E}} \:\mathrm{sin}\:\theta=\frac{{a}}{\mathrm{2}} \\ $$$$\Rightarrow{y}_{{E}} =\frac{{a}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{2}\:\mathrm{sin}\:\theta} \\ $$$${x}_{{F}} \:\mathrm{cos}\:\theta+\frac{{a}}{\mathrm{2}}\:\mathrm{sin}\:\theta=\frac{{a}}{\mathrm{2}} \\ $$$$\Rightarrow{x}_{{F}} =\frac{{a}\left(\mathrm{1}−\mathrm{sin}\:\theta\right)}{\mathrm{2}\:\mathrm{cos}\:\theta} \\ $$$$ \\ $$$${A}_{\Delta{BEF}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{a}}{\mathrm{2}}−\frac{{a}\left(\mathrm{1}−\mathrm{sin}\:\theta\right)}{\mathrm{2}\:\mathrm{cos}\:\theta}\right]\left[\frac{{a}}{\mathrm{2}}−\frac{{a}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{2}\:\mathrm{sin}\:\theta}\right] \\ $$$${A}_{\Delta{BEF}} =\frac{{a}^{\mathrm{2}} }{\mathrm{8}}\left[\frac{\mathrm{cos}\:\theta−\mathrm{1}+\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\right]\left[\frac{\mathrm{sin}\:\theta−\mathrm{1}+\mathrm{cos}\:\theta}{\:\mathrm{sin}\:\theta}\right] \\ $$$${A}_{\Delta{BEF}} =\frac{{a}^{\mathrm{2}} }{\mathrm{8}}×\frac{\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta} \\ $$$${A}_{\Delta{BEF}} =\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\left[\mathrm{1}−\frac{\mathrm{cos}\:\theta+\mathrm{sin}\:\theta−\mathrm{1}}{\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}\right] \\ $$$${A}_{{Green}} =\mathrm{4}{A}_{\Delta{BEF}} \\ $$$${A}_{{Green}} \left(\theta\right)={a}^{\mathrm{2}} \left[\mathrm{1}−\frac{\mathrm{cos}\:\theta+\mathrm{sin}\:\theta−\mathrm{1}}{\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}\right] \\ $$$${A}_{{Green}} \left(\theta\right)=\mathrm{2}{r}^{\mathrm{2}} \left[\mathrm{1}−\frac{\mathrm{cos}\:\theta+\mathrm{sin}\:\theta−\mathrm{1}}{\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}\right] \\ $$$${with}\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}} \\ $$$${A}_{{Green}} \left(\frac{\pi}{\mathrm{2}}+\theta\right)={A}_{{Green}} \left(\theta\right) \\ $$$$ \\ $$$${f}\left(\theta\right)=\mathrm{1}−\frac{\mathrm{cos}\:\theta+\mathrm{sin}\:\theta−\mathrm{1}}{\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta} \\ $$$${f}\left(\theta\right)=\mathrm{1}+\frac{\mathrm{2}\left[\mathrm{1}−\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{4}}\right)\right]}{\mathrm{sin}\:\mathrm{2}\theta} \\ $$$${max}.\:{f}\left(\theta\right)=\mathrm{1}+\frac{\mathrm{2}\left[\mathrm{1}−\sqrt{\mathrm{2}}\right]}{\mathrm{1}}=\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\approx\mathrm{0}.\mathrm{17} \\ $$$${at}\:\theta=\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow{max}.{A}_{{Green}} =\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right){a}^{\mathrm{2}} \approx\mathrm{0}.\mathrm{17}{a}^{\mathrm{2}} \approx\mathrm{0}.\mathrm{34}{r}^{\mathrm{2}} \\ $$