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


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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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