Find-maximum-value-of-x-16-y-2-y-4-x-2-for-all-real-x-and-y-

Question Number 128751 by bemath last updated on 10/Jan/21

Find maximum value of

∣ x ∣ \sqrt{16 - y^{2}} + ∣ y ∣ \sqrt{4 - x^{2}} for all real x and y .

Answered by mr W last updated on 10/Jan/21

a =∣ x ∣= 2 \sin α ⩾ 0, 0 ⩽ α ⩽ \frac{π}{2}

b =∣ y ∣= 4 \sin β ⩾ 0, 0 ⩽ β ⩽ \frac{π}{2}

a \sqrt{4^{2} - b^{2}} + b \sqrt{2^{2} - a^{2}}

= 8 \sin α \cos β + 8 \sin β \cos α

= 8 \sin (α + β)

m a x . = 8 a t \sin (α + β) = 1, e . g . α = β = \frac{π}{4}

Commented by liberty last updated on 10/Jan/21

i think not 8 is maximum value

Commented by mr W last updated on 10/Jan/21

i f t h e m a x i m u m f r o m 8 \sin (α + β)

i s n o t 8, t h e n i {d o n}^{'} t u n t e r s t a n d t h i s

w o r l d a n y m o r e :)

Commented by liberty last updated on 10/Jan/21

what is the reason by assuming ∣ x ∣ = 2 \sin α ?

Commented by mr W last updated on 10/Jan/21

s i n c e 4 - x^{2} ⩾ 0 \Rightarrow∣ x ∣⩽ 2

Commented by liberty last updated on 10/Jan/21

ok . you are right .

Answered by liberty last updated on 10/Jan/21

Using the inequality x y ⩽ \frac{x^{2} + y^{2}}{2}

we have {\begin{cases} ∣ x ∣ \sqrt{16 - y^{2}} ⩽ \frac{x^{2} + 16 - y^{2}}{2} \\ ∣ y ∣ \sqrt{4 - x^{2}} ⩽ \frac{y^{2} + 4 - x^{2}}{2} \end{cases}

adding the both equation we find

∣ x ∣ \sqrt{16 - y^{2}} + ∣ y ∣ \sqrt{4 - x^{2}} ⩽ \frac{x^{2} + 16 - y^{2} + y^{2} + 4 - x^{2}}{2} = 10

∴ its maximum value is 10

Commented by mr W last updated on 10/Jan/21

c a n y o u s a y w h a t v a l u e s x a n d y h a v e

s u c h t h a t y o u r e a c h t h e m a x i m u m

10 ?

Commented by liberty last updated on 10/Jan/21

10 is only an upper bound but cannot be

attained since the inequality xy ⩽ \frac{x^{2} + y^{2}}{2}

holds if and only if x = y, therefore the \max

value is attained ∣ x ∣ = \sqrt{16 - y^{2}} and ∣ y ∣= \sqrt{4 - x^{2}}

or x^{2} = 16 - y^{2} and y^{2} = 4 - x^{2} \Leftrightarrow x^{2} + y^{2} = 16 = 4

contradictory!

Commented by mr W last updated on 10/Jan/21

c o r r e c t!

s o b e c a r e f u l w h e n a d d i n g t w o o r

m o r e i n e q u a l i t i e s i f t h e y a r e n o t

i n d e p e n d e n t f r o m e a c h o t h e r!

Answered by mr W last updated on 10/Jan/21

a n o t h e r w a y :

f (a, b) = a \sqrt{16 - b^{2}} + b \sqrt{4 - a^{2}}

\frac{\partial f}{\partial a} = \sqrt{16 - b^{2}} - \frac{a b}{\sqrt{4 - a^{2}}} = 0

\frac{\partial f}{\partial b} = - \frac{a b}{\sqrt{16 - b^{2}}} + \sqrt{4 - a^{2}} = 0

\Rightarrow a b = \sqrt{(4 - a^{2}) (16 - b^{2})}

\Rightarrow 4 a^{2} + b^{2} = 16

\Rightarrow 16 - b^{2} = 4 a^{2}

\Rightarrow b = 2 \sqrt{4 - a^{2}}

\Rightarrow f_{m a x} = 2 a^{2} + 2 \sqrt{4 - a^{2}} \times \sqrt{4 - a^{2}} = 8

Commented by liberty last updated on 10/Jan/21

yes . nice

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