Let-k-be-a-real-number-such-that-the-inequality-x-3-6-x-k-has-a-solution-Find-the-maximum-value-of-k-

Question Number 110673 by Aina Samuel Temidayo last updated on 30/Aug/20

Let k be a real number such that the

inequality \sqrt{x - 3} + \sqrt{6 - x} ⩾ k has a

solution . Find the maximum value of

k .

Answered by Her_Majesty last updated on 30/Aug/20

f (x) = \sqrt{x - 3} + \sqrt{6 - x} \Rightarrow 3 ⩽ x ⩽ 6

m i n i m a a t t h e b o r d e r s

f (3) = f (6) = \sqrt{3} = k

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

Thanks but what do you mean by

minima at the borders ?

Commented by Her_Majesty last updated on 30/Aug/20

t h e b o r d e r s o f d e f i n i t i o n m y d e a r, o r d o

y o u s e e o t h e r r e l e v a n t b o r d e r s ?

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

Is minima at the borders true for

all functions ?

Commented by Her_Majesty last updated on 30/Aug/20

n o . b u t y o u a l w a y s h a v e t o c h e c k t h e b o r d e r s

h e r e :

y = \sqrt{x - 3} + \sqrt{6 - x}

y^{'} = \frac{1}{2 \sqrt{x - 3}} - \frac{1}{2 \sqrt{6 - x}} = \frac{\sqrt{6 - x} - \sqrt{x - 3}}{2 \sqrt{x - 3} \sqrt{6 - x}} = 0

\Leftrightarrow

\sqrt{6 - x} = \sqrt{x - 3}

\Leftrightarrow

6 - x = x - 3 \Rightarrow x = 9 / 2

i s t h e o n l y e x t r e m e b u t {i t}^{'} s t h e m a x i m u m

a t t h e b o r d e r s y^{'} i s u n d e f i n e d s o y o u c a n n o t

s o l v e t h e p r o b l e m i n t h e u s u a l w a y

Commented by Her_Majesty last updated on 30/Aug/20

s o r r y I m i s r e a d t h e q u e s t i o n .

i f {w e}^{'} r e l o o k i n g f o r a t l e a s t 1 s o l u t i o n o f

t h e i n e q u a l i t y w e i n d e e d g e t k = \sqrt{6} b e c a u s e

f (9 / 2) = \sqrt{6}

Answered by 1549442205PVT last updated on 30/Aug/20

Apply inequality AM - GM we have

{(\sqrt{x - 3} + \sqrt{6 - x})}^{2} ⩽ (1^{2} + 1^{2}) (x - 3 + 6 - x) = 6

\Rightarrow 0 < \sqrt{x - 3} + \sqrt{6 - x} ⩽ \sqrt{6} . (1) Hence, put

f (x) = \sqrt{x - 3} + \sqrt{6 - x}

Take k = \sqrt{6} then the inequality

\sqrt{x - 3} + \sqrt{6 - x} ⩾ k has at least one root

k = \sqrt{6} is greatest value because if

\exists k_{0} > \sqrt{6} such that

\sqrt{x - 3} + \sqrt{6 - x} ⩾ k_{0} then this inequality

has no root due to (1)