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Question-204433




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

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