If-x-2-y-2-xy-1-Find-range-of-E-x-3-y-xy-3-4-

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 \dots . ? ?

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! ��