if-x-2-xy-y-2-5-find-the-minimum-and-maximum-of-x-2-xy-y-2-

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 \dots

Answered by mr W last updated on 01/Oct/22

x^{2} + x y + y^{2} = c = 5 \dots (i)

l e t x^{2} - x y + y^{2} = k \dots (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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