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Question Number 53276 by ajfour last updated on 19/Jan/19

Commented by ajfour last updated on 19/Jan/19

$$Pinksemicirclemountsayellow$$$$equilateraltriangleofsidea.$$$$Findsidespandqofmaximum$$$$arearectangleinscribedwithin.$$

Answered by mr W last updated on 20/Jan/19

$$p=\frac{1}{2}\sqrt{a^2-q^2}+\frac{\sqrt{3}a}{2}(1-\frac{q}{a})$$$$A=pq=\frac{q}{2}[\sqrt{a^2-q^z}+\sqrt{3}a(1-\frac{q}{a})]$$$$A=\frac{a^2}{2}[\sqrt{1-{\lambda }^2}+\sqrt{3}(1-\lambda )]\lambda with\lambda =\frac{q}{a}$$$$\frac{d\left(\frac{2A}{a^2}\right)}{d\lambda }=0$$$$\sqrt{1-{\lambda }^2}+\sqrt{3}(1-\lambda )+\lambda [-\frac{\lambda }{\sqrt{1-{\lambda }^2}}-\sqrt{3}]=0$$$$\Rightarrow \sqrt{3(1-{\lambda }^2)}(1-2\lambda )=2{\lambda }^2-1$$$$\Rightarrow \lambda =0.6006$$$$\Rightarrow A_{max}=0.4478a^2$$

Commented by ajfour last updated on 20/Jan/19

$$ThanksSir,reallyprecise\mathrm{\&}excellent!$$

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