Question Number 154089 by iloveisrael last updated on 14/Sep/21 | ||
Answered by talminator2856791 last updated on 14/Sep/21 | ||
$$\: \\ $$$$\:\equiv\:\mathrm{max}\left(\frac{\mathrm{9}\centerdot\mathrm{sin}\left(\theta\right)\mathrm{9}\centerdot\mathrm{cos}\left(\theta\right)+\mathrm{3}\centerdot\mathrm{sin}\left(\theta\right)\mathrm{3}\centerdot\mathrm{cos}\left(\theta\right)}{\mathrm{2}}\right) \\ $$$$\:=\:\mathrm{max}\left(\frac{\mathrm{90}\left(\mathrm{sin}\left(\theta\right)\mathrm{cos}\left(\theta\right)\right)}{\mathrm{2}}\right) \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\mathrm{trigonometry}\:\mathrm{identity}: \\ $$$$\:\:\:\:\mathrm{sin}\left(\mathrm{2}\theta\right)=\:\mathrm{2}\centerdot\mathrm{sin}\left(\theta\right)\mathrm{cos}\left(\theta\right) \\ $$$$\: \\ $$$$\:\Rightarrow\:\mathrm{max}\left(\frac{\mathrm{45}\centerdot\mathrm{sin}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\right) \\ $$$$\:=\:\frac{\mathrm{45}}{\mathrm{2}} \\ $$$$\: \\ $$$$\:\mathrm{largest}\:\mathrm{area}: \\ $$$$\:\mathrm{2}×\frac{\mathrm{45}}{\mathrm{2}}+\mathrm{3}×\mathrm{9} \\ $$$$\:=\:\mathrm{72} \\ $$$$\: \\ $$ | ||
Answered by mr W last updated on 14/Sep/21 | ||
$${due}\:{to}\:{symmetry}: \\ $$$${a}_{{max}} =\frac{\mathrm{3}+\mathrm{9}}{\:\sqrt{\mathrm{2}}}=\mathrm{6}\sqrt{\mathrm{2}} \\ $$$${A}_{{max}} =\left(\mathrm{6}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{72} \\ $$$${at}\:\theta=\mathrm{45}° \\ $$ | ||