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


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

$If x^{2} + y^{2} = 9, then \max value of \frac{x^{3} + y^{3}}{x + y}$
Commented by bobhans last updated on 23/Jan/22

$all answer is great$
	Answered by mahdipoor last updated on 23/Jan/22

	$A = \frac{(x + y) (x^{2} + y^{2} - x y)}{x + y} = 9 - x y (x \neq - y)$ $= 9 \pm x \sqrt{9 - x^{2}}$ $\frac{d A}{d x} = 0 = \pm (\frac{- 2 x^{2}}{2 \sqrt{9 - x^{2}}} + \sqrt{9 - x^{2}}) \Rightarrow x = \pm \sqrt{\frac{9}{2}}$ $\Rightarrow A (\pm \sqrt{\frac{9}{2}}) = 9 \pm \frac{9}{2} \Rightarrow A_{m a x} = 9 + \frac{9}{2} = 13.5$
	Answered by mr W last updated on 23/Jan/22

	$\sqrt{a b} ⩽ \frac{a + b}{2}$ $\frac{x^{3} + y^{3}}{x + y} = x^{2} + y^{2} - x y = x^{2} + y^{2} + \sqrt{{(- x)}^{2} y^{2}}$ $⩽ x^{2} + y^{2} + \frac{{(- x)}^{2} + y^{2}}{2} = 9 + \frac{9}{2} = \frac{27}{2}$ $i . e . {(\frac{x^{3} + y^{3}}{x + y})}_{m a x} = \frac{27}{2}$
	Answered by FelipeLz last updated on 23/Jan/22
	$x^2 +y^2 = 9 → { ((x = 3cos(t))),((y = 3sin(t))) :} v = ((x^3 +y^3 )/(x+y)) = (((x+y)(x^2 −xy+y^2 ))/(x+y)) = x^2 −xy+y^2 = (9/2)[2−sin(2t)] sin(2t) ∈ [−1, 1] ∀t 2t = sin^(−1) (−1) 2t = ((3π)/2) t = ((3π)/4) → v = ((27)/2)$
	$x^{2} + y^{2} = 9 \to {\begin{cases} x = 3 \cos (t) \\ y = 3 \sin (t) \end{cases}$ $v = \frac{x^{3} + y^{3}}{x + y} = \frac{(x + y) (x^{2} - x y + y^{2})}{x + y} = x^{2} - x y + y^{2} = \frac{9}{2} [2 - \sin (2 t)]$ $\sin (2 t) \in [- 1, 1] \forall t$ $2 t = \sin^{- 1} (- 1)$ $2 t = \frac{3 π}{2}$ $t = \frac{3 π}{4} \to v = \frac{27}{2}$
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