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


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
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 by mr 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 by Shrinava 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 by Shrinava last updated on 18/Nov/22

	$Thank you so much dear professor$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum