Question Number 145449 by solihin last updated on 05/Jul/21
Answered by Olaf_Thorendsen last updated on 05/Jul/21
$${p}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\pi{a}+\mathrm{2}{a}+\mathrm{2}{b}\:=\:\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}\right){a}+\mathrm{2}{b}\:=\:\mathrm{5} \\ $$$${A}\:=\:\frac{\pi{a}^{\mathrm{2}} }{\mathrm{8}}+{ab} \\ $$$${A}\:=\:\frac{\pi{a}^{\mathrm{2}} }{\mathrm{8}}+{a}×\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}−\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}\right){a}\right) \\ $$$${A}\:=\:−\left(\frac{\pi}{\mathrm{8}}+\mathrm{1}\right){a}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{2}}{a} \\ $$$${A}'\left({a}\right)\:=\:−\left(\frac{\pi}{\mathrm{4}}+\mathrm{2}\right){a}+\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${maximum}\:{for}\:{A}'\left({a}\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{a}\:=\:\frac{\mathrm{5}}{\mathrm{2}\left(\frac{\pi}{\mathrm{4}}+\mathrm{2}\right)}\:=\:\frac{\mathrm{10}}{\pi+\mathrm{8}} \\ $$$${b}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}−\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}\right){a}\right) \\ $$$${b}\:=\:\left(\frac{\mathrm{6}−\pi}{\mathrm{4}}\right)\frac{\mathrm{10}}{\pi+\mathrm{8}}\:=\:\frac{\mathrm{5}\left(\mathrm{6}−\pi\right)}{\mathrm{2}\left(\mathrm{8}+\pi\right)} \\ $$$${A}\:=\:\frac{\pi{a}^{\mathrm{2}} }{\mathrm{8}}+{ab} \\ $$$${A}\:=\:\frac{\mathrm{100}\pi}{\mathrm{8}\left(\pi+\mathrm{8}\right)^{\mathrm{2}} }+\frac{\mathrm{10}}{\pi+\mathrm{8}}.\frac{\mathrm{5}\left(\mathrm{6}−\pi\right)}{\mathrm{2}\left(\mathrm{8}+\pi\right)} \\ $$$${A}\:=\:\frac{\mathrm{25}\pi+\mathrm{50}\left(\mathrm{6}−\pi\right)}{\mathrm{2}\left(\pi+\mathrm{8}\right)^{\mathrm{2}} } \\ $$$${A}\:=\:\frac{\mathrm{25}\left(\mathrm{12}−\pi\right)}{\mathrm{2}\left(\pi+\mathrm{8}\right)^{\mathrm{2}} } \\ $$