if-a-b-c-d-e-8-and-a-2-b-2-c-2-d-2-e-2-16-what-is-the-maximal-value-of-a-

Question Number 180641 by mr W last updated on 14/Nov/22

i f a + b + c + d + e = 8 a n d

a^{2} + b^{2} + c^{2} + d^{2} + e^{2} = 16, w h a t i s t h e

m a x i m a l v a l u e o f a ?

Answered by Emrice last updated on 14/Nov/22

a < 4

Commented by Rasheed.Sindhi last updated on 15/Nov/22

a ⩽ 4

Commented by mr W last updated on 15/Nov/22

i f a = 4, t h e n b = c = d = e = 0, a n d

a + b + c + d + e = 4 \neq 8, t h e r e f o r e a m u s t

b e l e s s t h a n 4 .

Commented by Rasheed.Sindhi last updated on 15/Nov/22

Y e s s i r I u n d e r s t o o d .

Answered by a.lgnaoui last updated on 15/Nov/22

4

Answered by manxsol last updated on 15/Nov/22

s i a = 4

f_{1} 4 + \dots \dots . = 8 Σ (b + c + d + e) = 4

f_{2} > 16

a \neq 4

Commented by manxsol last updated on 15/Nov/22

s i a = 4

f_{1} 4 + \dots \dots . = 8 Σ (b + c + d + e) = 4

f_{2} > 16

a \neq 4

\begin{array}{ccc} a & b & c & d & e \\ 3 & 2 & 1 & 1 & 1 & 8 \\ 9 & 4 & 1 & 1 & 1 & 16 \end{array}

Commented by mr W last updated on 15/Nov/22

a \neq 4, {t h a t}^{'} s t r u e . b u t a < 4 {d o e s n}^{'} t

m e a n t h e n e x t v a l u e f o r a i s 3, b e c a u s e

a, b, c, d, e {m u s n}^{'} t b e i n t e g e r .

Answered by mr W last updated on 15/Nov/22

t o g e t a a s l a r g e a s p o s s i b l e, t h e o t h e r

4 n u m b e r s s h o u l d b e a s c l o s e t o e a c h

o t h e r a s p o s s i b l e . t h a t m e a n s t h e

o t h e r n u m b e r s s h o u l d b e e q u a l, s a y

e q u a l t o x . t h e n w e h a v e

a + 4 x = 8

a^{2} + 4 x^{2} = 16

\Rightarrow 4 a^{2} + {(8 - a)}^{2} = 64

5 a^{2} - 16 a = 0

\Rightarrow a = \frac{16}{5} = 3.2

\Rightarrow x = \frac{6}{5} = 1.2

t h a t m e a n s w e g e t a_{m a x} = 3.2

w h e n b = c = d = e = 1.2 .

Commented by manxsol last updated on 15/Nov/22

Thanks Mr.W. I am learn.

Commented by mr W last updated on 15/Nov/22

a n o t h e r m e t h o d

(l i t t l e t h i n k i n g, j u s t a p p l y i n g f o r m u l a)

a = 8 - (b + c + d + e)

{(8 - b - c - d - e)}^{2} + b^{2} + c^{2} + d^{2} + e^{2} - 16 = 0

Φ = 8 - (b + c + d + e) + λ [{(8 - b - c - d - e)}^{2} + b^{2} + c^{2} + d^{2} + e^{2} - 16]

\frac{\partial Φ}{\partial b} = - 1 + λ [- 2 (8 - b - c - d - e) + 2 b] = 0

\Rightarrow b = \frac{1}{2 λ} + 8 - (b + c + d + e)

s i m i l a r l y

\Rightarrow c = \frac{1}{2 λ} + 8 - (b + c + d + e)

\Rightarrow d = \frac{1}{2 λ} + 8 - (b + c + d + e)

\Rightarrow e = \frac{1}{2 λ} + 8 - (b + c + d + e)

\Rightarrow b = c = d = e = x, s a y

\frac{\partial Φ}{\partial λ} = {(8 - b - c - d - e)}^{2} + b^{2} + c^{2} + d^{2} + e^{2} - 16 = 0

\Rightarrow {(8 - 4 x)}^{2} + 4 x^{2} - 16 = 0

5 x^{2} - 16 x + 12 = 0

\Rightarrow x = \frac{8 \pm 2}{5} = 2 o r \frac{6}{5}

\Rightarrow a = 0 (m i n) o r \frac{16}{5} (m a x)

Commented by mr W last updated on 16/Nov/22

U s i n g L a g r a n g e M u l t i p l i e r m e t h o d

f o r f i n d i n g l o c a l m a x i m u m o r

m i n i m u m o f a f u n c t i o n w i t h m o r e

t h a n o n e v a r i a b l e u n d e r a c o n s t r a i n t .

Commented by manolex last updated on 16/Nov/22

w h a t t e o r y f o r m a t i o n Φ ?

Commented by mr W last updated on 16/Nov/22

https://www.wikihow.com/Use-Lagrange-Multipliers