Let-x-y-z-be-nonnegative-real-numbers-which-satisfy-x-y-z-1-Find-minimum-value-of-Q-2-x-2-y-2-z-

Question Number 119790 by bemath last updated on 27/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, w h i c h s a t i s f y x + y + z = 1

F i n d m i n i m u m v a l u e o f

Q = \sqrt{2 - x} + \sqrt{2 - y} + \sqrt{2 - z} .

Answered by 1549442205PVT last updated on 27/Oct/20

Put x + y = 2 a, x + z = 2 b, y + z = 2 c

\Rightarrow a + b + c = x + y + z = 1; a, b, c ⩾ 0

Q = \sqrt{1 + y + z} + \sqrt{1 + x + z} + \sqrt{1 + x + y}

= \sqrt{1 + 2 a} + \sqrt{1 + 2 b} + \sqrt{1 + 2 c} with a + b + c = 1

Q (a, b) = \sqrt{1 + 2 a} + \sqrt{1 + 2 b} + \sqrt{3 - 2 (a + b)}

We find extremum of

{\begin{cases} \frac{\partial Q}{\partial a} = 0 \\ \frac{\partial Q}{\partial b} = 0 \end{cases}

\Leftrightarrow {\begin{cases} \frac{1}{\sqrt{1 + 2 a}} - \frac{1}{\sqrt{3 - 2 (a + b)}} = 0 (1) \\ \frac{1}{\sqrt{1 + 2 b}} - \frac{1}{\sqrt{3 - 2 (a + b)}} = 0 (2) \end{cases}

\Rightarrow \frac{1}{\sqrt{1 + 2 a}} = \frac{1}{\sqrt{1 + 2 b}} \Rightarrow a = b

Replace into (1) we get

\sqrt{3 - 4 a} = \sqrt{1 + 2 a} \Leftrightarrow 3 - 4 a = 1 + 2 a

\Leftrightarrow 6 a = 2 \Rightarrow a = b = 1 / 3

A = \frac{\partial^{2} Q}{\partial a^{2}} ∣_{(\frac{1}{3}, \frac{1}{3})} = \frac{- 1}{(1 + 2 a) \sqrt{1 + 2 a}} - \frac{1}{[3 - 2 (a + b)] \sqrt{3 - 2 (a + b)}} ∣_{(\frac{1}{3}, \frac{1}{3})}

= \frac{- 9}{5 \sqrt{15}} - \frac{9}{5 \sqrt{15}} = \frac{- 18}{5 \sqrt{15}} < 0

C = \frac{\partial^{2} Q}{\partial b^{2}} ∣_{(\frac{1}{3}, \frac{1}{3})} = \frac{- 1}{(1 + 2 b) \sqrt{1 + 2 b}} - \frac{1}{[3 - 2 (a + b)] \sqrt{3 - 2 (a + b)}} ∣_{(\frac{1}{3}, \frac{1}{3})}

= \frac{- 18}{\sqrt{515}} < 0

\frac{\partial^{2} Q}{\partial a \partial b} ∣_{(\frac{1}{3}, \frac{1}{3})} = \frac{\partial Q}{\partial b} (\frac{1}{\sqrt{1 + 2 a}} - \frac{1}{\sqrt{3 - 2 (a + b)}})

B = \frac{1}{\sqrt{3 - 2 (a + b)}} ∣_{(\frac{1}{3}, \frac{1}{3})} = \frac{3}{\sqrt{15}}

Δ = AC - B^{2} = {[\frac{18}{5 \sqrt{15}}]}^{2} - \frac{9}{15} = \frac{324 - 225}{25.15} > 0

Since A < 0, Q has maximum

Q_{\max} = Q (\frac{1}{3}, \frac{1}{3}) = \sqrt{15} when (x, y, z)

= (\frac{1}{3}, \frac{1}{3}, \frac{1}{3})

On the other hand, consider Q (x, y, z) at

the bounded points we have

Q (0, 1, 0) = Q (0, 1, 0) = Q (0, 0, 1) =

= 1 + 2 \sqrt{2} = 3.828 . . < \sqrt{15} \approx 3.873

Q (\frac{1}{2}, \frac{1}{2}, 0) = \sqrt{2} + \sqrt{6} \approx 3.8637 > 1 + 2 \sqrt{2}

Comparing the above values we see

Tthe smallest value of Q equal to 1 + 2 \sqrt{2}

when (x, y, z) = (0, 0, 1) and all the its

permutation

Answered by ebi last updated on 27/Oct/20

Q = {(2 - x)}^{\frac{1}{2}} + {(2 - y)}^{\frac{1}{2}} + {(2 - z)}^{\frac{1}{2}}

c o n s t r a i n t, x + y + z = 1 = F

t h e s o l u t i o n l i e s o n 0 ⩽ x, y, z ⩽ 1

▽ Q = λ ▽ F

⟨ Q_{x}, Q_{y}, Q_{z} ⟩ = λ ⟨ F_{x}, F_{y}, F_{z} ⟩

λ = l a g r a n g e m u l t i p l i e r

Q_{x} = λ F_{x}

- \frac{1}{2 \sqrt{2 - x}} = λ

{(- \frac{1}{2 \sqrt{2 - x}})}^{2} = λ^{2}

λ^{2} = \frac{1}{4 (2 - x)}

x = 2 - \frac{1}{4 λ^{2}} \dots \dots (1)

Q_{y} = λ F_{y}

- \frac{1}{2 \sqrt{2 - y}} = λ

λ^{2} = \frac{1}{4 (2 - y)}

y = 2 - \frac{1}{4 λ^{2}} \dots \dots (2)

Q_{z} = λ F_{z}

- \frac{1}{2 \sqrt{2 - z}} = λ

λ^{2} = \frac{1}{4 (2 - z)}

z = 2 - \frac{1}{4 λ^{2}} \dots \dots (3)

x + y + z = 1 \dots \dots (4)

s u b s t i t u t e (1), (2), (3) i n t o (4)

(2 - \frac{1}{4 λ^{2}}) + (2 - \frac{1}{4 λ^{2}}) + (2 - \frac{1}{4 λ^{2}}) = 1

- \frac{3}{4 λ^{2}} = - 5

λ^{2} = \frac{3}{20}

x = 2 - \frac{1}{4 (\frac{3}{20})} = \frac{1}{3}

y = 2 - \frac{1}{4 (\frac{3}{20})} = \frac{1}{3}

z = 2 - \frac{1}{4 (\frac{3}{20})} = \frac{1}{3}

∴ t h e s o l u t i o n i s (\frac{1}{3}, \frac{1}{3}, \frac{1}{3})

G (\frac{1}{3}, \frac{1}{3}, \frac{1}{3}) = \sqrt{2 - \frac{1}{3}} + \sqrt{2 - \frac{1}{3}} + \sqrt{2 - \frac{1}{3}}

= 3 \sqrt{\frac{5}{3}} = \sqrt{15} \approx 3.873

b e c a u s e o f t h i s p o i n t i s t h e o n l y

s t a t i o n a r y p o i n t, w e h a v e t o c o n s i d e r

o t h e r p o i n t s f o r m a x i m a o r m i n i m a

c o n s i d e r t h e s e 3 p o i n t s :

(1, 0, 0), (0, 1, 0), (0, 0, 1)

G (1, 0, 0) = G (0, 1, 0) = G (0, 0, 1)

= 1 + 2 \sqrt{2} \approx 3.828

t h u s,

G (\frac{1}{3}, \frac{1}{3}, \frac{1}{3}) = 3.873 \to m a x i m u m

G (1, 0, 0) = G (0, 1, 0) = G (0, 0, 1) = 3.828 \to m i n i m u m

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