Let x,y,z be non negative real numbers such that x+y+z=1. Find the extremum of F = 2x^2 +y+3z^2 .


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Question Number 119724 by benjo_mathlover last updated on 26/Oct/20

$L e t x, y, z b e n o n n e g a t i v e r e a l n u m b e r s$ $s u c h t h a t x + y + z = 1 . F i n d t h e e x t r e m u m$ $o f F = 2 x^{2} + y + 3 z^{2} .$
	Answered by mindispower last updated on 26/Oct/20

	$y = 1 - x - z$ $Missing \left or extra \right$ $⩾ \frac{19}{24}$
	Commented by bemath last updated on 26/Oct/20

	$w h a t t h e m a x v a l u e s i r ?$
	Answered by 1549442205PVT last updated on 27/Oct/20

	$F = {2 x}^{2} + {3 z}^{2} + 1 - x - z = 2 {(x - \frac{1}{4})}^{2}$ $Missing \left or extra \right$ $\Rightarrow F_{\min} = \frac{19}{24} when (x, y, z) = (\frac{1}{4}, \frac{7}{12}, \frac{1}{6})$ $b) Since x + y + z = 1, x, y, z ⩾ 0 (hypothesis)$ $\Rightarrow 0 ⩽ x, y, z ⩽ 1 \Rightarrow x^{2} ⩽ x, y^{2} ⩽ y, z^{2} ⩽ z$ $\Rightarrow F = {2 x}^{2} + y + {3 z}^{2} ⩽ {3 x}^{2} + y + {3 z}^{2} ⩽$ $3 x + 3 y + 3 z = 3 (x + y + z) = 3 . The equality$ $ocurrs if and only if x = y = 0, z = 1$ $Hence F_{\max} = 3 when (x, y, z) = (0, 0, 1)$
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