If x^2 +y^2 +xy=1. Find range of E = x^3 y+xy^3 +4 .


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Question Number 46478 by rahul 19 last updated on 27/Oct/18

$I f x^{2} + y^{2} + x y = 1 . F i n d r a n g e o f$ $E = x^{3} y + {x y}^{3} + 4 .$
Commented by rahul 19 last updated on 27/Oct/18

$S i r, a n s . i s$ $2 ⩽ E ⩽ \frac{38}{9} .$ $I s i t w r o n g a s i c a n n o t f i n d a n y m i s t a k e$ $i n y o u r s o l u t i o n . . . . ? ?$
	Answered by MJS last updated on 27/Oct/18

	$x^{2} + y x + y^{2} - 1 = 0$ $this is an ellipse with center (\begin{matrix} 0 \\ 0 \end{matrix}) and θ = - 45 °$ $y = - \frac{x}{2} \pm \frac{\sqrt{4 - 3 x^{2}}}{2} \Rightarrow - \frac{2 \sqrt{3}}{3} ⩽ x ⩽ \frac{2 \sqrt{3}}{3}$ $\Rightarrow - \frac{2 \sqrt{3}}{3} ⩽ y ⩽ \frac{2 \sqrt{3}}{3} [because of symmetry]$ $E (x) = x^{3} (- \frac{x}{2} \pm \frac{\sqrt{4 - 3 x^{2}}}{2}) + x {(- \frac{x}{2} \pm \frac{\sqrt{4 - 3 x^{2}}}{2})}^{3} + 4 =$ $= \frac{x^{4}}{2} - \frac{3 x^{2}}{2} + 4 \pm \frac{x (x^{2} + 1) \sqrt{4 - 3 x^{2}}}{2}$ $solving E^{'} (x) = 0 we get minima and maxima$ $at x = \pm 1 and x = \pm \frac{\sqrt{3}}{3} \Rightarrow range (E) = [2; \frac{38}{9}]$
	Commented by rahul 19 last updated on 27/Oct/18
	thanks sir! ��
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