Question-204433

Question Number 204433 by emilagazade last updated on 17/Feb/24

Commented by emilagazade last updated on 17/Feb/24

f i n d m a x v a l u e f o r ∣ D B ∣ + ∣ B C ∣

Answered by deleteduser1 last updated on 17/Feb/24

L e t ∠ C A D = θ \Rightarrow {C D}^{2} = 3^{2} + 64 \times 2 - 2 \times 3 \times 8 \sqrt{2} c o s (θ)

\Rightarrow C D = \sqrt{137 - 48 \sqrt{2} c o s (θ)}

{A C}^{2} = {A B}^{2} + {B C}^{2} \Rightarrow 128 = {(3 + B D)}^{2} + {B C}^{2}

\Rightarrow 119 = {B D}^{2} + {B C}^{2} + 6 B D

{C D}^{2} = {B D}^{2} + {B C}^{2} \Rightarrow 137 - 48 \sqrt{2} c o s (θ) = 119 - 6 B D

\Rightarrow B D = \frac{48 \sqrt{2} c o s (θ) - 18}{6} = 8 \sqrt{2} c o s (θ) - 3

\Rightarrow B C = \sqrt{128 - 128 {c o s}^{2} θ} = 8 \sqrt{2} s i n (θ)

\Rightarrow B D + D C = 8 \sqrt{2} (s i n θ + c o s θ) - 3

⩽ 8 \sqrt{2} (\sqrt{1 + 1}) ({s i n}^{2} θ + {c o s}^{2} θ) - 3 = 13

E q u a l i t y w h e n s i n (θ) = c o s (θ) \Rightarrow θ = 45 ° .

Answered by mr W last updated on 17/Feb/24

Commented by mr W last updated on 17/Feb/24

{(∣ D B ∣ + ∣ B C ∣)}_{m a x} m e a n s a l s o

{(∣ A B ∣ + ∣ B C ∣)}_{m a x}, w h i c h i s

∣ {A B}^{'} ∣ + ∣ B^{'} C ∣= 2 ∣ {A B}^{'} ∣= 2 \times \frac{8 \sqrt{2}}{\sqrt{2}} = 16 .

\Rightarrow {(∣ D B ∣ + ∣ B C ∣)}_{m a x} = 16 - 3 = 13 ✓

Commented by mr W last updated on 17/Feb/24

M e t h o d I I

s a y ∠ C A B = θ

B C = 8 \sqrt{2} \sin θ

D B = 8 \sqrt{2} \cos θ - 3

S = D B + B C = 8 \sqrt{2} (\cos θ + \sin θ) - 3

= 16 \sin (\frac{π}{4} + θ) - 3

S_{m a x} = 16 - 3 = 13 a t θ = \frac{π}{4} .

Commented by emilagazade last updated on 17/Feb/24

n i c e t h a n k y o u a l o t

Answered by cortano12 last updated on 18/Feb/24

\underset{⏟}{}