if-x-y-z-0-find-the-maximum-of-x-2y-3z-x-2-y-2-z-2-

Question Number 181318 by mr W last updated on 23/Nov/22

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

\frac{∣ x + 2 y + 3 z ∣}{\sqrt{x^{2} + y^{2} + z^{2}}} .

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

s e e Q 181875 f o r o t h e r s o l u t i o n s

Answered by Frix last updated on 24/Nov/22

\frac{∣ x + 2 y + 3 z ∣}{\sqrt{x^{2} + y^{2} + z^{2}}}

z = - (x + y)

\frac{\sqrt{2} ∣ 2 x + y ∣}{2 \sqrt{x^{2} + x y + y^{2}}}

y = c x

\frac{\sqrt{2} ∣ c + 2 ∣}{2 \sqrt{c^{2} + c + 1}}

\Rightarrow \min at c = - 2

\frac{\sqrt{2}}{2} \times \frac{d [\frac{∣ c + 2 ∣}{\sqrt{c^{2} + c + 1}}]}{d c} = - \frac{3 \sqrt{2} c (c + 2)}{4 ∣ c + 2 ∣ \sqrt{{(c^{2} + c + 1)}^{3}}} = 0

\Rightarrow \max at c = 0

\Rightarrow 0 ⩽ \frac{\sqrt{2} ∣ c + 2 ∣}{2 \sqrt{c^{2} + c + 1}} ⩽ \sqrt{2}

\forall r \in R ∖ {0} :

the minimum is at (x, y, z) = (r, - 2 r, r)

the maximum is at (x, y, z) = (r, 0, - r)

Answered by a.lgnaoui last updated on 24/Nov/22

x + 2 y + 3 z = 2 (x + y + z) + z - x = z - x

∣ x + 2 y + 3 z ∣<∣ x ∣ + ∣ z ∣

x^{2} + y^{2} + z^{2} = {(x + y + z)}^{2} - 2 [y (x + z) + x z]

= 2 [{(x + z)}^{2} - x z]

\frac{∣ x + 2 y + 3 z ∣}{\sqrt{x^{2} + y^{2} + z^{2}}} < \frac{∣ x + z ∣}{∣ x + z ∣ \sqrt{2 - \frac{2 x z}{{(x + z)}^{2}}}}

< \frac{∣ x + z ∣}{\sqrt{\frac{2 {(x + z)}^{2}}{1}}} < \frac{1}{\sqrt{2}}

t h e m a x i m u m i s \frac{\sqrt{2}}{2}

Commented by Frix last updated on 24/Nov/22

wrong!

with x = 1 \land y = 0 \land z = - 1

\frac{∣ x + 2 y + 3 z ∣}{\sqrt{x^{2} + y^{2} + z^{2}}} = \frac{∣ 1 - 3 ∣}{\sqrt{1 + 0 + 1}} = \frac{2}{\sqrt{2}} = \sqrt{2} > \frac{\sqrt{2}}{2}

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