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Question Number 145449 by solihin last updated on 05/Jul/21

Answered by Olaf_Thorendsen last updated on 05/Jul/21

$$p=\frac{1}{2}\pi a+2a+2b=(\frac{\pi }{2}+2)a+2b=5$$$$A=\frac{\pi a^2}{8}+ab$$$$A=\frac{\pi a^2}{8}+a\times \frac{1}{2}(5-(\frac{\pi }{2}+2)a)$$$$A=-(\frac{\pi }{8}+1)a^2+\frac{5}{2}a$$$$A^{\prime }\left(a\right)=-(\frac{\pi }{4}+2)a+\frac{5}{2}$$$$maximumfor{A}^{\prime }\left(a\right)=0$$$$\Rightarrow a=\frac{5}{2(\frac{\pi }{4}+2)}=\frac{10}{\pi +8}$$$$b=\frac{1}{2}(5-(\frac{\pi }{2}+2)a)$$$$b=\left(\frac{6-\pi }{4}\right)\frac{10}{\pi +8}=\frac{5(6-\pi )}{2(8+\pi )}$$$$A=\frac{\pi a^2}{8}+ab$$$$A=\frac{100\pi }{8{(\pi +8)}^2}+\frac{10}{\pi +8}.\frac{5(6-\pi )}{2(8+\pi )}$$$$A=\frac{25\pi +50(6-\pi )}{2{(\pi +8)}^2}$$$$A=\frac{25(12-\pi )}{2{(\pi +8)}^2}$$

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