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Question Number 164912 by bobhans last updated on 23/Jan/22

If x^2 +y^2 = 9 , then max value of ((x^3 +y^3 )/(x+y))

$$\:\:\mathrm{If}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{9}\:,\:\mathrm{then}\:\mathrm{max}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} }{\mathrm{x}+\mathrm{y}} \\ $$

Commented by bobhans last updated on 23/Jan/22

all answer is great

$$\mathrm{all}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{great} \\ $$

Answered by mahdipoor last updated on 23/Jan/22

A=(((x+y)(x^2 +y^2 −xy))/(x+y))=9−xy (x≠−y) =9±x(√(9−x^2 )) (dA/dx)=0=±(((−2x^2 )/(2(√(9−x^2 ))))+(√(9−x^2 )))⇒x=±(√(9/2)) ⇒A(±(√(9/2)))=9±(9/2)⇒A_(max) =9+(9/2)=13.5

$${A}=\frac{\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{xy}\right)}{{x}+{y}}=\mathrm{9}−{xy}\:\:\:\left({x}\neq−{y}\right) \\ $$$$=\mathrm{9}\pm{x}\sqrt{\mathrm{9}−{x}^{\mathrm{2}} } \\ $$$$\frac{{dA}}{{dx}}=\mathrm{0}=\pm\left(\frac{−\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}+\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }\right)\Rightarrow{x}=\pm\sqrt{\frac{\mathrm{9}}{\mathrm{2}}} \\ $$$$\Rightarrow{A}\left(\pm\sqrt{\frac{\mathrm{9}}{\mathrm{2}}}\right)=\mathrm{9}\pm\frac{\mathrm{9}}{\mathrm{2}}\Rightarrow{A}_{{max}} =\mathrm{9}+\frac{\mathrm{9}}{\mathrm{2}}=\mathrm{13}.\mathrm{5} \\ $$

Answered by mr W last updated on 23/Jan/22

(√(ab))≤((a+b)/2) ((x^3 +y^3 )/(x+y))=x^2 +y^2 −xy=x^2 +y^2 +(√((−x)^2 y^2 )) ≤x^2 +y^2 +(((−x)^2 +y^2 )/2)=9+(9/2)=((27)/2) i.e. (((x^3 +y^3 )/(x+y)))_(max) =((27)/2)

$$\sqrt{{ab}}\leqslant\frac{{a}+{b}}{\mathrm{2}} \\ $$$$\frac{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} }{{x}+{y}}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{xy}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\sqrt{\left(−{x}\right)^{\mathrm{2}} {y}^{\mathrm{2}} } \\ $$$$\leqslant{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\frac{\left(−{x}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{9}+\frac{\mathrm{9}}{\mathrm{2}}=\frac{\mathrm{27}}{\mathrm{2}} \\ $$$${i}.{e}.\:\left(\frac{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} }{{x}+{y}}\right)_{{max}} =\frac{\mathrm{27}}{\mathrm{2}} \\ $$

Answered by FelipeLz last updated on 23/Jan/22

x^2 +y^2 = 9 → { ((x = 3cos(t))),((y = 3sin(t))) :} v = ((x^3 +y^3 )/(x+y)) = (((x+y)(x^2 −xy+y^2 ))/(x+y)) = x^2 −xy+y^2 = (9/2)[2−sin(2t)] sin(2t) ∈ [−1, 1] ∀t 2t = sin^(−1) (−1) 2t = ((3π)/2) t = ((3π)/4) → v = ((27)/2)

$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:=\:\mathrm{9}\:\rightarrow\:\begin{cases}{{x}\:=\:\mathrm{3cos}\left({t}\right)}\\{{y}\:=\:\mathrm{3sin}\left({t}\right)}\end{cases} \\ $$$${v}\:=\:\frac{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} }{{x}+{y}}\:=\:\frac{\left({x}+{y}\right)\left({x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} \right)}{{x}+{y}}\:=\:{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} \:=\:\frac{\mathrm{9}}{\mathrm{2}}\left[\mathrm{2}−\mathrm{sin}\left(\mathrm{2}{t}\right)\right] \\ $$$$\mathrm{sin}\left(\mathrm{2}{t}\right)\:\in\:\left[−\mathrm{1},\:\mathrm{1}\right]\:\:\forall{t} \\ $$$$\mathrm{2}{t}\:=\:\mathrm{sin}^{−\mathrm{1}} \left(−\mathrm{1}\right) \\ $$$$\mathrm{2}{t}\:=\:\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$${t}\:=\:\frac{\mathrm{3}\pi}{\mathrm{4}}\:\rightarrow\:{v}\:=\:\frac{\mathrm{27}}{\mathrm{2}} \\ $$

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