Question Number 164912 by bobhans last updated on 23/Jan/22
$$\:\:\mathrm{If}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{9}\:,\:\mathrm{then}\:\mathrm{max}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} }{\mathrm{x}+\mathrm{y}} \\ $$
Commented by bobhans last updated on 23/Jan/22
$$\mathrm{all}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{great} \\ $$
Answered by mahdipoor last updated on 23/Jan/22
$${A}=\frac{\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{xy}\right)}{{x}+{y}}=\mathrm{9}−{xy}\:\:\:\left({x}\neq−{y}\right) \\ $$$$=\mathrm{9}\pm{x}\sqrt{\mathrm{9}−{x}^{\mathrm{2}} } \\ $$$$\frac{{dA}}{{dx}}=\mathrm{0}=\pm\left(\frac{−\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}+\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }\right)\Rightarrow{x}=\pm\sqrt{\frac{\mathrm{9}}{\mathrm{2}}} \\ $$$$\Rightarrow{A}\left(\pm\sqrt{\frac{\mathrm{9}}{\mathrm{2}}}\right)=\mathrm{9}\pm\frac{\mathrm{9}}{\mathrm{2}}\Rightarrow{A}_{{max}} =\mathrm{9}+\frac{\mathrm{9}}{\mathrm{2}}=\mathrm{13}.\mathrm{5} \\ $$
Answered by mr W last updated on 23/Jan/22
$$\sqrt{{ab}}\leqslant\frac{{a}+{b}}{\mathrm{2}} \\ $$$$\frac{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} }{{x}+{y}}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{xy}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\sqrt{\left(−{x}\right)^{\mathrm{2}} {y}^{\mathrm{2}} } \\ $$$$\leqslant{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\frac{\left(−{x}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{9}+\frac{\mathrm{9}}{\mathrm{2}}=\frac{\mathrm{27}}{\mathrm{2}} \\ $$$${i}.{e}.\:\left(\frac{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} }{{x}+{y}}\right)_{{max}} =\frac{\mathrm{27}}{\mathrm{2}} \\ $$
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)](https://www.tinkutara.com/question/Q164917.png)
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:=\:\mathrm{9}\:\rightarrow\:\begin{cases}{{x}\:=\:\mathrm{3cos}\left({t}\right)}\\{{y}\:=\:\mathrm{3sin}\left({t}\right)}\end{cases} \\ $$$${v}\:=\:\frac{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} }{{x}+{y}}\:=\:\frac{\left({x}+{y}\right)\left({x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} \right)}{{x}+{y}}\:=\:{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} \:=\:\frac{\mathrm{9}}{\mathrm{2}}\left[\mathrm{2}−\mathrm{sin}\left(\mathrm{2}{t}\right)\right] \\ $$$$\mathrm{sin}\left(\mathrm{2}{t}\right)\:\in\:\left[−\mathrm{1},\:\mathrm{1}\right]\:\:\forall{t} \\ $$$$\mathrm{2}{t}\:=\:\mathrm{sin}^{−\mathrm{1}} \left(−\mathrm{1}\right) \\ $$$$\mathrm{2}{t}\:=\:\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$${t}\:=\:\frac{\mathrm{3}\pi}{\mathrm{4}}\:\rightarrow\:{v}\:=\:\frac{\mathrm{27}}{\mathrm{2}} \\ $$