Max. value

Algebra Level 4

2 ( a − x ) [ x + x 2 + b 2 ] 2(a-x) \left [ x + \sqrt{ x^2 + b^2 } \right ]

If x x is real, then state the maximum value of the expression above in terms of a a and b b .

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a b \frac {a}{b} a 2 + b 2 { a }^{ 2 }+{ b }^{ 2 } a b ab a 2 { a }^{ 2 }

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Utkarsh Bansal by Utkarsh Bansal , 23, India

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1 solution

Utkarsh Bansal
Feb 24, 2015

L e t t = x + x 2 + b 2 ⇒ 1 t = 1 x + x 2 + b 2 = x 2 + b 2 − x b 2 ∴ t − b 2 t = 2 x a n d t + b 2 t = 2 x 2 + b 2 T h u s 2 ( a − x ) [ x + x 2 + b 2 ] = [ 2 a − t + b 2 t ] . ( t ) = 2 a t − t 2 + b 2 = a 2 + b 2 − ( a 2 − 2 a t + t 2 ) = a 2 + b 2 − ( a − t ) 2 T h e r e f o r e y = 2 ( a − x ) [ x + x 2 + b 2 ] ≤ a 2 + b 2 h e n c e m a x . v a l u e o f y i s a 2 + b 2 Let\quad t=x+\sqrt { { x }^{ 2 }+{ b }^{ 2 } } \\ \Rightarrow \frac { 1 }{ t } =\frac { 1 }{ x+\sqrt { { x }^{ 2 }+{ b }^{ 2 } } } =\frac { \sqrt { { x }^{ 2 }+{ b }^{ 2 } } -x }{ { b }^{ 2 } } \\ \therefore \quad t-\frac { { b }^{ 2 } }{ t } =2x\quad and\quad t+\frac { { b }^{ 2 } }{ t } =2\sqrt { { x }^{ 2 }+{ b }^{ 2 } } \\ Thus\quad 2\left( a-x \right) \left[ x+\sqrt { { x }^{ 2 }+{ b }^{ 2 } } \right] =\left[ 2a-t+\frac { { b }^{ 2 } }{ t } \right] .\left( t \right) \\ =2at-{ t }^{ 2 }+{ b }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }-\left( { a }^{ 2 }-2at+{ t }^{ 2 } \right) \\ ={ a }^{ 2 }+{ b }^{ 2 }-{ \left( a-t \right) }^{ 2 }\\ Therefore\quad y=2\left( a-x \right) \left[ x+\sqrt { { x }^{ 2 }+{ b }^{ 2 } } \right] \le { a }^{ 2 }+{ b }^{ 2 }\\ hence\quad max.\quad value\quad of\quad y\quad is\quad \boxed { { a }^{ 2 }+{ b }^{ 2 } }

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theres a mistake. 't' is not equal to (b^2/t). please check carefully. (3rd line)

Chirag Singapore - 6 years, 3 months ago

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thanks for pointing out the mistake.I've corrected it.

Utkarsh Bansal - 6 years, 3 months ago

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