if x^2 +xy+y^2 =5, find the minimum and maximum of x^2 −xy+y^2 .


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Question Number 177130 by mr W last updated on 01/Oct/22

$i f x^{2} + x y + y^{2} = 5, f i n d t h e m i n i m u m$ $a n d m a x i m u m o f x^{2} - x y + y^{2} .$
Commented by Frix last updated on 01/Oct/22

$I think$ $\frac{5}{3} ⩽ x^{2} - x y + y^{2} ⩽ 15$
Commented by Frix last updated on 01/Oct/22

$x^{2} \pm x y + y^{2} = c$ $\Rightarrow$ $\frac{c}{3} ⩽ x^{2} \mp x y + y^{2} ⩽ 3 c$
Commented by mr W last updated on 01/Oct/22

$t h a n k s s i r!$
Commented by Frix last updated on 01/Oct/22

$you are right, I {don}^{'} t know what I did before . . .$
	Answered by mr W last updated on 01/Oct/22

	$x^{2} + x y + y^{2} = c = 5 . . . (i)$ $l e t x^{2} - x y + y^{2} = k . . . (i i)$ $[(i) - (i i)] / 2 :$ $\Rightarrow x y = \frac{c - k}{2} \Rightarrow y = \frac{c - k}{2 x}$ $x^{2} + \frac{c - k}{2} + {(\frac{c - k}{2 x})}^{2} = c$ $x^{4} - (\frac{c + k}{2}) x^{2} + {(\frac{c - k}{2})}^{2} = 0$ $Δ = {(\frac{c + k}{2})}^{2} - 4 {(\frac{c - k}{2})}^{2} ⩾ 0$ $(k - 3 c) (3 k - c) ⩽ 0$ $\Rightarrow \frac{c}{3} ⩽ k ⩽ 3 c$ $i . e . \frac{c}{3} ⩽ x^{2} - x y + y^{2} ⩽ 3 c$ $w i t h c = 5 :$ $m i n i m u m = \frac{5}{3}$ $m a x i m u m = 3 \times 5 = 15$
	Commented by Strengthenchen last updated on 06/Oct/22

	$h o w c a n i t o k n o w u s e t h i s w a y t o g e t c o r r e c t i n g a n s w e r i f i n e v e r s e e m t h i s f o r m q u e s t i o n b e f o r e ?$
	Answered by mr W last updated on 02/Oct/22

	$a n o t h e r m e t h o d$ $l e t x = u + v, y = u - v$ $x^{2} + x y + y^{2} = c$ ${(u + v)}^{2} + (u + v) (u - v) + {(u - v)}^{2} = c$ $3 u^{2} + v^{2} = c \to e l i p s e$ $\Rightarrow u = \sqrt{\frac{c}{3}} \cos θ, v = \sqrt{c} \sin θ$ $k = x^{2} - x y + y^{2}$ $= {(u + v)}^{2} - (u + v) (u - v) + {(u - v)}^{2}$ $= u^{2} + 3 v^{2}$ $= \frac{c \cos^{2} θ}{3} + 3 c \sin^{2} θ$ $= \frac{c \cos^{2} θ}{3} + 3 c (1 - \cos^{2} θ)$ $= (3 - \frac{8 \cos^{2} θ}{3}) c$ $\Rightarrow k_{m i n} = (3 - \frac{8}{3}) c = \frac{c}{3}$ $\Rightarrow k_{m a x} = (3 - 0) c = 3 c$
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