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Question Number 153901 by mr W last updated on 11/Sep/21

Commented by mr W last updated on 12/Sep/21

$$1)find{\left(ab\right)}_{max}=?$$$$2)provethat{S}_1=S_2.$$

Answered by EDWIN88 last updated on 12/Sep/21

Commented by mr W last updated on 12/Sep/21

$$thanksalotsir!$$

Commented by Tawa11 last updated on 12/Sep/21

$$\mathrm{Weldone}\mathrm{sir}.$$

Answered by mr W last updated on 12/Sep/21

Commented by mr W last updated on 12/Sep/21

$$\left(1\right)$$$$\alpha +\beta =45$$$$\frac{\tan \alpha +\tan \alpha }{1-\tan \alpha \tan \beta }=1$$$$\frac{1}{\tan \alpha }+\frac{1}{\tan \beta }=\frac{1}{\tan \alpha }\times \frac{1}{\tan \beta }-1$$$$\Rightarrow \frac{1}{\tan \beta }=\frac{\frac{1}{\tan \alpha }+1}{\frac{1}{\tan \alpha }-1}$$$$$$$$\frac{\sin (\alpha +45)}{\sin \alpha }=\frac{l}{a}$$$$\frac{1}{\tan \alpha }+1=\frac{\sqrt{2}l}{a}...\left(i\right)$$$$similarly$$$$\frac{1}{\tan \beta }+1=\frac{\sqrt{2}l}{b}...\left(ii\right)$$$$\frac{2l^2}{ab}=(\frac{1}{\tan \alpha }+1)(\frac{1}{\tan \beta }+1)$$$$=2(\frac{1}{\tan \alpha }+1)\left(\frac{\frac{1}{\tan \alpha }}{\frac{1}{\tan \alpha }-1}\right)$$$$lett=\frac{1}{\tan \alpha }$$$$\Rightarrow P=\frac{l^2}{ab}=\frac{t(t+1)}{t-1}$$$$\frac{dP}{dt}=\frac{2t+1}{t-1}-\frac{t(t+1)}{{(t-1)}^2}=0$$$$t^2-2t-1=0$$$$\Rightarrow t=\frac{1}{\tan \alpha }=1+\sqrt{2}$$$$\Rightarrow \tan \alpha =\sqrt{2}-1\Rightarrow \alpha =22.5°=\beta$$$$\Rightarrow \tan \beta =\sqrt{2}-1$$$$\frac{2l^2}{{\left(ab\right)}_{max}}={(2+\sqrt{2})}^2$$$$\Rightarrow {\left(ab\right)}_{max}={(\sqrt{2}-1)}^2{l}^2\approx 0.171l^2$$

Commented by mr W last updated on 12/Sep/21

$$\left(2\right)$$$$\frac{L}{c}=\frac{\sin (45+\alpha )}{\sin 45}=\frac{\sin (45+45-\beta )}{\sin 45}=\sqrt{2}\cos \beta$$$$\frac{L}{d}=\frac{\sin (45+\beta )}{\sin 45}=\frac{\sin (45+45-\alpha )}{\sin 45}=\sqrt{2}\cos \alpha$$$$\frac{L^2}{cd}=2\cos \alpha \cos \beta$$$$\Rightarrow cd=\frac{L^2}{2\cos \alpha \cos \beta }$$$$e=\frac{L}{\cos \alpha }$$$$f=\frac{L}{\cos \beta }$$$$\Rightarrow ef=\frac{L^2}{\cos \alpha \cos \beta }=2cd$$$$S_1=\frac{cd\sin 45°}{2}$$$$S_1+S_2=\frac{ef\sin 45°}{2}=2S_1$$$$\Rightarrow S_1=S_2$$

Commented by liberty last updated on 12/Sep/21

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