If a,b,c<0 and abc(a+b+c)=64 Then find min of P=2a+b+c


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Question Number 180873 by Shrinava last updated on 18/Nov/22

$If a, b, c < 0 and abc (a + b + c) = 64$ $Then find \min of P = 2 a + b + c$
Commented bymr W last updated on 18/Nov/22

$p l e a s e c h e c k t h e q u e s t i o n . s i n c e$ $a, b, c < 0, P h a s m a x i m u m, n o t m i n .$ $o r y o u m e a n a, b, c > 0 ?$
Commented byShrinava last updated on 18/Nov/22

$sorry dear professor > 0$
	Answered by mr W last updated on 18/Nov/22

	$d u e t o s y m m e t r y, P h a s e x t r e m u m$ $w h e n b = c = k, s a y$ $P = 2 a + b + c = 2 (a + k)$ ${a k}^{2} (a + 2 k) = 64$ $k^{2} a^{2} + 2 k^{3} a - 64 = 0$ $\Rightarrow a = \sqrt{k^{2} + \frac{64}{k^{2}}} - k$ $\Rightarrow P = 2 \sqrt{k^{2} + \frac{64}{k^{2}}} ⩾ 2 \sqrt{2 \times 8} = 8 = P_{m i n}$
	Commented byShrinava last updated on 18/Nov/22

	$Thank you so much dear professor$
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