If-x-2-y-2-9-then-max-value-of-x-3-y-3-x-y-

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 \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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