if-a-b-c-d-e-f-10-and-a-2-b-2-c-2-d-2-e-2-f-2-25-find-a-min-and-f-max-

Question Number 205914 by mr W last updated on 02/Apr/24

i f a + b + c + d + e + f = 10 a n d

a^{2} + b^{2} + c^{2} + d^{2} + e^{2} + f^{2} = 25, f i n d

a_{m i n} a n d f_{m a x} .

Commented by Tinku Tara last updated on 03/Apr/24

Question :

Why is \min = - 5 and \max = + 5

not the solution ?

Commented by mr W last updated on 03/Apr/24

i f m i n = - 5, s a y a = - 5, t o f u l f i l l

a^{2} + b^{2} + c^{2} + d^{2} + e^{2} + f^{2} = 25, w e m u s t

h a v e b = c = d = e = f = 0, i . e .

a + b + c + d + e + f = - 5, b u t t h i s i s

c o n t r a d i c t i o n t o a + b + c + d + e + f = 10,

t h e r e f o r e a \neq - 5 .

Answered by mr W last updated on 05/Apr/24

\underset{―}{M e t h o d I I}

a = 10 - (b + c + d + e + f)

{(10 - (b + c + d + e + f))}^{2} + b^{2} + c^{2} + d^{2} + e^{2} + f^{2} - 25 = 0

F = a = 10 - (b + c + d + e + f) + λ [{(10 - (b + c + d + e + f))}^{2} + b^{2} + c^{2} + d^{2} + e^{2} + f^{2} - 25]

\frac{\partial F}{\partial b} = - 1 + λ [- 2 (10 - (b + c + d + e + f)) + 2 b] = 0

=> - \frac{1}{2} + λ [- 10 + (b + c + d + e + f) + b] = 0

\Rightarrow b = \frac{1}{2} + 10 λ - λ (b + c + d + e + f)

s i m i l a r l y

\Rightarrow c = d = e = f = \frac{1}{2} + 10 λ - λ (b + c + d + e + f)

i . e . f o r a_{m i n} o r a_{m a x},

b = c = d = e = f = t, s a y

t h e n w e g e t a s s h o w n a b o v e

a_{m i n} = \frac{10 - 5 \sqrt{10}}{6}, a_{m a x} = \frac{10 + 5 \sqrt{10}}{6}

Answered by mr W last updated on 05/Apr/24

\underset{―}{M e t h o d I I I}

\vec{p} = (a, b, c, d, e, f)

\vec{q} = (1, 1, 1, 1, 1, 1)

\cos θ = \frac{\vec{p} \cdot \vec{q}}{∣ p ∣∣ q ∣}

= \frac{a + b + c + d + e + f}{\sqrt{6 (a^{2} + b^{2} + c^{2} + d^{2} + e^{2} + f^{2}})} = \frac{2}{\sqrt{6}}

a_{m i n / m a x} = 5 \cos (α \pm θ)

w i t h α = \cos^{- 1} \frac{1}{\sqrt{6}}

a_{m i n / m a x} = 5 \times (\frac{1}{\sqrt{6}} \times \frac{2}{\sqrt{6}} \mp \frac{\sqrt{5}}{\sqrt{6}} \times \frac{\sqrt{2}}{\sqrt{6}})

= \frac{5 (2 \mp \sqrt{10})}{6} ✓

Commented by mr W last updated on 05/Apr/24

Answered by mr W last updated on 04/Apr/24

\underset{―}{M e t h o d I}

i f b + c + d + e + f = s,

b^{2} + c^{2} + d^{2} + e^{2} + f^{2} ⩾ \frac{{(b + c + d + e + f)}^{2}}{5} = \frac{s^{2}}{5}

e q u a l i t y h o l d s w h e n b = c = d = e = f = \frac{s}{5}

s i n c e a + b + c + d + e + f = 10 a n d

a^{2} + b^{2} + c^{2} + d^{2} + e^{2} + f^{2} = 25, s u c h t h a t

a i s a s l a r g e a s p o s s i b l e, b + c + d + e + f

s h o u l d b e a s s m a l l a s p o s s i b l e a n d

b e s i d e s b, c, d, e, f s h o u l d b e e q u a l .

s a y b = c = d = e = f = t, t h e n

a + 5 t = 10 \Rightarrow t = \frac{10 - a}{5}

a^{2} + 5 t^{2} = 25 \Rightarrow a^{2} + 5 \times {(\frac{10 - a}{5})}^{2} = 25

6 a^{2} - 20 a - 25 = 0

a = \frac{10 \pm 5 \sqrt{10}}{6}

i . e . a_{m i n} = \frac{10 - 5 \sqrt{10}}{6}, a_{m a x} = \frac{10 + 5 \sqrt{10}}{6}

Answered by mr W last updated on 04/Apr/24

\underset{―}{M e t h o d I V}

f (x) = {(x - b)}^{2} + {(x - c)}^{2} + {(x - d)}^{2} + {(x - e)}^{2} + {(x - f)}^{2} ⩾ 0

f (x) = 5 x^{2} - 2 (b + c + d + e + f) + (b^{2} + c^{2} + d^{2} + e^{2} + f^{2}) ⩾ 0

f (x) = 5 x^{2} - 2 (10 - a) x + (25 - a^{2}) ⩾ 0

\Rightarrow {(10 - a)}^{2} - 5 (25 - a^{2}) ⩽ 0

\Rightarrow 6 a^{2} - 20 a - 25 ⩽ 0

\Rightarrow \frac{10 - 5 \sqrt{10}}{6} ⩽ a ⩽ \frac{10 + 5 \sqrt{10}}{6}

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