Question Number 204433 by emilagazade last updated on 17/Feb/24
Commented by emilagazade last updated on 17/Feb/24
$${find}\:{max}\:{value}\:{for}\:\mid{DB}\mid+\mid{BC}\mid \\ $$
Answered by deleteduser1 last updated on 17/Feb/24
$${Let}\:\angle{CAD}=\theta\Rightarrow{CD}^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} +\mathrm{64}×\mathrm{2}−\mathrm{2}×\mathrm{3}×\mathrm{8}\sqrt{\mathrm{2}}{cos}\left(\theta\right) \\ $$$$\Rightarrow{CD}=\sqrt{\mathrm{137}−\mathrm{48}\sqrt{\mathrm{2}}{cos}\left(\theta\right)} \\ $$$${AC}^{\mathrm{2}} ={AB}^{\mathrm{2}} +{BC}^{\mathrm{2}} \Rightarrow\mathrm{128}=\left(\mathrm{3}+{BD}\right)^{\mathrm{2}} +{BC}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{119}={BD}^{\mathrm{2}} +{BC}^{\mathrm{2}} +\mathrm{6}{BD} \\ $$$${CD}^{\mathrm{2}} ={BD}^{\mathrm{2}} +{BC}^{\mathrm{2}} \Rightarrow\mathrm{137}−\mathrm{48}\sqrt{\mathrm{2}}{cos}\left(\theta\right)=\mathrm{119}−\mathrm{6}{BD} \\ $$$$\Rightarrow{BD}=\frac{\mathrm{48}\sqrt{\mathrm{2}}{cos}\left(\theta\right)−\mathrm{18}}{\mathrm{6}}=\mathrm{8}\sqrt{\mathrm{2}}{cos}\left(\theta\right)−\mathrm{3} \\ $$$$\Rightarrow{BC}=\sqrt{\mathrm{128}−\mathrm{128}{cos}^{\mathrm{2}} \theta}=\mathrm{8}\sqrt{\mathrm{2}}{sin}\left(\theta\right) \\ $$$$\Rightarrow{BD}+{DC}=\mathrm{8}\sqrt{\mathrm{2}}\left({sin}\theta+{cos}\theta\right)−\mathrm{3} \\ $$$$\leqslant\mathrm{8}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{1}+\mathrm{1}}\right)\left({sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta\right)−\mathrm{3}=\mathrm{13} \\ $$$${Equality}\:{when}\:{sin}\left(\theta\right)={cos}\left(\theta\right)\Rightarrow\theta=\mathrm{45}°. \\ $$
Answered by mr W last updated on 17/Feb/24
Commented by mr W last updated on 17/Feb/24
$$\left(\mid{DB}\mid+\mid{BC}\mid\right)_{{max}} \:{means}\:{also}\: \\ $$$$\left(\mid{AB}\mid+\mid{BC}\mid\right)_{{max}} ,\:{which}\:{is}\: \\ $$$$\mid{AB}'\mid+\mid{B}'{C}\mid=\mathrm{2}\mid{AB}'\mid=\mathrm{2}×\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}}=\mathrm{16}. \\ $$$$\Rightarrow\left(\mid{DB}\mid+\mid{BC}\mid\right)_{{max}} =\mathrm{16}−\mathrm{3}=\mathrm{13}\:\checkmark \\ $$
Commented by mr W last updated on 17/Feb/24
$${Method}\:{II} \\ $$$${say}\:\angle{CAB}=\theta \\ $$$${BC}=\mathrm{8}\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta \\ $$$${DB}=\mathrm{8}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\theta−\mathrm{3} \\ $$$${S}={DB}+{BC}=\mathrm{8}\sqrt{\mathrm{2}}\:\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)−\mathrm{3} \\ $$$$\:\:\:=\mathrm{16}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+\theta\right)−\mathrm{3} \\ $$$${S}_{{max}} =\mathrm{16}−\mathrm{3}=\mathrm{13}\:{at}\:\theta=\frac{\pi}{\mathrm{4}}. \\ $$
Commented by emilagazade last updated on 17/Feb/24
$${nice}\:{thank}\:{you}\:{a}\:{lot} \\ $$
Answered by cortano12 last updated on 18/Feb/24
$$\:\underbrace{\:} \\ $$