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-

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)