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

Find maximum value of ∣x∣ (√(16−y^2 )) + ∣y∣ (√(4−x^2 )) for all real x and y.

$$\mathrm{Find}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\mid\mathrm{x}\mid\:\sqrt{\mathrm{16}−\mathrm{y}^{\mathrm{2}} }\:+\:\mid\mathrm{y}\mid\:\sqrt{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }\:\mathrm{for}\:\mathrm{all}\:\mathrm{real}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}. \\ $$$$ \\ $$

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

a=∣x∣=2 sin α ≥0, 0≤α≤(π/2) b=∣y∣=4 sin β ≥0, 0≤β≤(π/2) a(√(4^2 −b^2 ))+b(√(2^2 −a^2 )) =8 sin α cos β+8 sin β cos α =8 sin (α+β) max. =8 at sin (α+β)=1, e.g. α=β=(π/4)

$${a}=\mid{x}\mid=\mathrm{2}\:\mathrm{sin}\:\alpha\:\geqslant\mathrm{0},\:\mathrm{0}\leqslant\alpha\leqslant\frac{\pi}{\mathrm{2}} \\ $$$${b}=\mid{y}\mid=\mathrm{4}\:\mathrm{sin}\:\beta\:\geqslant\mathrm{0},\:\mathrm{0}\leqslant\beta\leqslant\frac{\pi}{\mathrm{2}} \\ $$$${a}\sqrt{\mathrm{4}^{\mathrm{2}} −{b}^{\mathrm{2}} }+{b}\sqrt{\mathrm{2}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$=\mathrm{8}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta+\mathrm{8}\:\mathrm{sin}\:\beta\:\mathrm{cos}\:\alpha \\ $$$$=\mathrm{8}\:\mathrm{sin}\:\left(\alpha+\beta\right) \\ $$$${max}.\:=\mathrm{8}\:{at}\:\mathrm{sin}\:\left(\alpha+\beta\right)=\mathrm{1},\:{e}.{g}.\:\alpha=\beta=\frac{\pi}{\mathrm{4}} \\ $$

Commented by liberty last updated on 10/Jan/21

i think not 8 is maximum value

$$\mathrm{i}\:\mathrm{think}\:\mathrm{not}\:\mathrm{8}\:\mathrm{is}\:\mathrm{maximum}\:\mathrm{value} \\ $$

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

if the maximum from 8 sin (α+β) is not 8, then i don′t unterstand this world any more :)

$${if}\:{the}\:{maximum}\:{from}\:\mathrm{8}\:\mathrm{sin}\:\left(\alpha+\beta\right) \\ $$$${is}\:{not}\:\mathrm{8},\:{then}\:{i}\:{don}'{t}\:{unterstand}\:{this}\: \\ $$$$\left.{world}\:{any}\:{more}\::\right) \\ $$

Commented by liberty last updated on 10/Jan/21

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

$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{reason}\:\mathrm{by}\:\mathrm{assuming}\mid\mathrm{x}\mid\:=\:\mathrm{2}\:\mathrm{sin}\:\alpha\:?\: \\ $$

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

since 4−x^2 ≥0 ⇒∣x∣≤2

$${since}\:\mathrm{4}−{x}^{\mathrm{2}} \geqslant\mathrm{0}\:\Rightarrow\mid{x}\mid\leqslant\mathrm{2} \\ $$

Commented by liberty last updated on 10/Jan/21

ok. you are right.

$$\mathrm{ok}.\:\mathrm{you}\:\mathrm{are}\:\mathrm{right}. \\ $$

Answered by liberty last updated on 10/Jan/21

Using the inequality xy ≤ ((x^2 +y^2 )/2) we have { ((∣x∣ (√(16−y^2 )) ≤ ((x^2 +16−y^2 )/2))),((∣y∣ (√(4−x^2 )) ≤ ((y^2 +4−x^2 )/2))) :} adding the both equation we find ∣x∣ (√(16−y^2 )) + ∣y∣ (√(4−x^2 )) ≤ ((x^2 +16−y^2 +y^2 +4−x^2 )/2) = 10 ∴ its maximum value is 10

$$\:\mathrm{Using}\:\mathrm{the}\:\mathrm{inequality}\:\:{xy}\:\leqslant\:\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\mathrm{we}\:\mathrm{have}\:\begin{cases}{\mid\mathrm{x}\mid\:\sqrt{\mathrm{16}−\mathrm{y}^{\mathrm{2}} }\:\leqslant\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{16}−\mathrm{y}^{\mathrm{2}} }{\mathrm{2}}}\\{\mid\mathrm{y}\mid\:\sqrt{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }\:\leqslant\:\frac{\mathrm{y}^{\mathrm{2}} +\mathrm{4}−\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{adding}\:\mathrm{the}\:\mathrm{both}\:\mathrm{equation}\:\mathrm{we}\:\mathrm{find}\: \\ $$$$\:\mid\mathrm{x}\mid\:\sqrt{\mathrm{16}−\mathrm{y}^{\mathrm{2}} }\:+\:\mid\mathrm{y}\mid\:\sqrt{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }\:\leqslant\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{16}−\mathrm{y}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{4}−\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:=\:\mathrm{10} \\ $$$$\therefore\:\mathrm{its}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{is}\:\mathrm{10}\: \\ $$$$ \\ $$

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

can you say what values x and y have such that you reach the maximum 10?

$${can}\:{you}\:{say}\:{what}\:{values}\:\:{x}\:{and}\:{y}\:{have} \\ $$$${such}\:{that}\:{you}\:{reach}\:{the}\:{maximum} \\ $$$$\mathrm{10}? \\ $$

Commented by liberty last updated on 10/Jan/21

10 is only an upper bound but cannot be attained since the inequality xy ≤ ((x^2 +y^2 )/2) holds if and only if x=y , therefore the max value is attained ∣x∣ =(√(16−y^2 )) and ∣y∣=(√(4−x^2 )) or x^2 =16−y^2 and y^2 =4−x^2 ⇔ x^2 +y^2 = 16=4 contradictory !

$$\mathrm{10}\:\mathrm{is}\:\mathrm{only}\:\mathrm{an}\:\mathrm{upper}\:\mathrm{bound}\:\mathrm{but}\:\mathrm{cannot}\:\mathrm{be}\: \\ $$$$\mathrm{attained}\:\mathrm{since}\:\mathrm{the}\:\mathrm{inequality}\:\mathrm{xy}\:\leqslant\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{holds}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if}\:\mathrm{x}=\mathrm{y}\:,\:\mathrm{therefore}\:\mathrm{the}\:\mathrm{max} \\ $$$$\mathrm{value}\:\mathrm{is}\:\mathrm{attained}\:\mid\mathrm{x}\mid\:=\sqrt{\mathrm{16}−\mathrm{y}^{\mathrm{2}} }\:\mathrm{and}\:\mid\mathrm{y}\mid=\sqrt{\mathrm{4}−\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{or}\:\mathrm{x}^{\mathrm{2}} =\mathrm{16}−\mathrm{y}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{y}^{\mathrm{2}} =\mathrm{4}−\mathrm{x}^{\mathrm{2}} \:\Leftrightarrow\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\:\mathrm{16}=\mathrm{4} \\ $$$$\mathrm{contradictory}\:! \\ $$

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

correct! so be careful when adding two or more inequalities if they are not independent from each other!

$${correct}! \\ $$$${so}\:{be}\:{careful}\:{when}\:{adding}\:{two}\:{or} \\ $$$${more}\:{inequalities}\:{if}\:{they}\:{are}\:{not} \\ $$$${independent}\:{from}\:{each}\:{other}! \\ $$

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

an other way: f(a,b)=a(√(16−b^2 ))+b(√(4−a^2 )) (∂f/∂a)=(√(16−b^2 ))−((ab)/( (√(4−a^2 ))))=0 (∂f/∂b)=−((ab)/( (√(16−b^2 ))))+(√(4−a^2 ))=0 ⇒ab=(√((4−a^2 )(16−b^2 ))) ⇒4a^2 +b^2 =16 ⇒16−b^2 =4a^2 ⇒b=2(√(4−a^2 )) ⇒f_(max) =2a^2 +2(√(4−a^2 ))×(√(4−a^2 ))=8

$${an}\:{other}\:{way}: \\ $$$${f}\left({a},{b}\right)={a}\sqrt{\mathrm{16}−{b}^{\mathrm{2}} }+{b}\sqrt{\mathrm{4}−{a}^{\mathrm{2}} } \\ $$$$\frac{\partial{f}}{\partial{a}}=\sqrt{\mathrm{16}−{b}^{\mathrm{2}} }−\frac{{ab}}{\:\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}=\mathrm{0} \\ $$$$\frac{\partial{f}}{\partial{b}}=−\frac{{ab}}{\:\sqrt{\mathrm{16}−{b}^{\mathrm{2}} }}+\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow{ab}=\sqrt{\left(\mathrm{4}−{a}^{\mathrm{2}} \right)\left(\mathrm{16}−{b}^{\mathrm{2}} \right)} \\ $$$$\Rightarrow\mathrm{4}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{16} \\ $$$$\Rightarrow\mathrm{16}−{b}^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} \\ $$$$\Rightarrow{b}=\mathrm{2}\sqrt{\mathrm{4}−{a}^{\mathrm{2}} } \\ $$$$\Rightarrow{f}_{{max}} =\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }×\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }=\mathrm{8} \\ $$

Commented by liberty last updated on 10/Jan/21

yes. nice

$$\mathrm{yes}.\:\mathrm{nice} \\ $$

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