if-a-b-c-0-find-the-maximum-of-a-2b-3c-a-2-b-2-c-2-

Question Number 181875 by mr W last updated on 01/Dec/22

i f a + b + c = 0, f i n d t h e m a x i m u m o f

\frac{∣ a + 2 b + 3 c ∣}{\sqrt{a^{2} + b^{2} + c^{2}}} .

Commented by mr W last updated on 01/Dec/22

Q 181318 r e p o s t e d, a s k i n g f o r m o r e

s o l u t i o n s .

Commented by SEKRET last updated on 01/Dec/22

what happened to previos solutions

Commented by mr W last updated on 01/Dec/22

t h e y a r e t h e r e, n o t h i n g h a p p e n e d .

Answered by SEKRET last updated on 01/Dec/22

\max (\sqrt{2})

Commented by mr W last updated on 01/Dec/22

p l e a s e s h o w y o u r s o l u t i o n .

Answered by SEKRET last updated on 01/Dec/22

c = - (b + a) a + 2 b - 3 (b + a) = - (b + 2 a)

a^{2} + b^{2} + {(a + b)}^{2} = 2 a^{2} + 2 b^{2} + 2 ab

\frac{∣ b + 2 a ∣}{\sqrt{2} \cdot \sqrt{a^{2} + b^{2} + ab}} b = ax

\frac{∣ a ∣ \cdot ∣ x + 2 ∣}{\sqrt{2} \cdot ∣ a ∣ \cdot \sqrt{x^{2} + x + 1}} = \frac{∣ x + 2 ∣}{\sqrt{2} \cdot \sqrt{x^{2} + x + 1}} = f (x)

f^{'} (x) = 0 \frac{- 3 x}{2 \sqrt{2} \cdot \sqrt{{(x^{2} + x + 1)}^{3}}} = 0

x = 0 f (0) = \frac{2}{\sqrt{2}} = \sqrt{2}

Commented by mr W last updated on 01/Dec/22

t h a n k s!

Answered by mr W last updated on 02/Dec/22

Missing \left or extra \right

b = - (a + c)

f = \frac{∣ a + 2 b + 3 c ∣}{\sqrt{a^{2} + b^{2} + c^{2}}}

= \frac{∣ a - 2 a - 2 c + 3 c ∣}{\sqrt{a^{2} + {(a + c)}^{2} + c^{2}}}

= \frac{∣ c - a ∣}{\sqrt{2 (a^{2} + a c + c^{2})}}

= \frac{∣ k - 1 ∣}{\sqrt{2 (1 + k + k^{2})}} w i t h k = \frac{c}{a}

= \frac{∣ k - 1 ∣}{\sqrt{2 ({(k - 1)}^{2} + 3 (k - 1) + 3)}}

= \frac{∣ u ∣}{\sqrt{2 (u^{2} + 3 u + 3)}} w i t h u = k - 1

= \frac{1}{\sqrt{2 (\frac{3}{u^{2}} + \frac{3}{u} + 1)}}

= \frac{1}{\sqrt{6 [{(\frac{1}{u} + \frac{1}{2})}^{2} + \frac{1}{12}]}}

⩽ \frac{1}{\sqrt{6 \times \frac{1}{12}}} = \sqrt{2}

\Rightarrow m a x i m u m = \sqrt{2}

a t u = - 2 = k - 1 = \frac{c}{a} - 1 \Rightarrow c + a = 0, b = 0

Commented by SEKRET last updated on 01/Dec/22

super

Answered by mr W last updated on 02/Dec/22

Missing \left or extra \right

i m a g e w e h a v e a p l a n e w i t h f o l l o w i n g

e q u a t i o n :

a x + b y + c z = 0

t h i s p l a n e p a s s e s t h r o u g h t h e o r i g i n

A (0, 0, 0) .

s i n c e a + b + c = 0, i t m e a n s t h e p l a n e

p a s s e s a l s o t h r o u g h t h e p o i n t B (1, 1, 1) .

w e k n o w \frac{∣ a + 2 b + 3 c ∣}{\sqrt{a^{2} + b^{2} + c^{2}}} r e p r e s e n t s t h e

d i s t a n c e f r o m t h e p o i n t P (1, 2, 3) t o t h e

p l a n e a x + b y + c z = 0 .

n o w w e o n l y n e e d t o f i n d t h e l a r g e s t

d i s t a n c e f r o m t h e p o i n t P (1, 2, 3) t o t h e

p l a n e a x + b y + c z = 0 .

s i n c e t h e p l a n e p a s s e s t h r o u g h t h e

p o i n t s A (0, 0, 0) a n d B (1, 1, 1), t h e

l a r g e s t d i s t a n c e f r o m p o i n t P t o t h e

p l a n e i s t h e d i s t a n c e f r o m p o i n t P t o

t h e l i n e A B .

A B = (1, 1, 1)

A P = (1, 2, 3)

A P \times A B = (1, 2, 3) \times (1, 1, 1) = (1, 2, 1)

d = \frac{∣ A P \times A B ∣}{∣ A B ∣} = \frac{\sqrt{1^{2} + 2^{2} + 1^{2}}}{\sqrt{1^{2} + 1^{2} + 1^{2}}} = \frac{\sqrt{6}}{\sqrt{3}} = \sqrt{2}

t h a t m e a n s m a x i m u m o f \frac{∣ a + 2 b + 3 c ∣}{\sqrt{a^{2} + b^{2} + c^{2}}}

i s \sqrt{2} .

Commented by mr W last updated on 01/Dec/22

Commented by mr W last updated on 02/Dec/22

a l t e r n a t i v e :

\cos θ = \frac{(1, 2, 3) \cdot (1, 1, 1)}{\sqrt{(1^{2} + 1^{2} + 1^{2}) (1^{2} + 2^{2} + 3^{2})}} = \frac{\sqrt{6}}{\sqrt{7}}

\sin θ = \frac{1}{\sqrt{7}}

d = A P \sin θ = \sqrt{14} \times \frac{1}{\sqrt{7}} = \sqrt{2}

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