Find max value of x^2 −3xy−2y^2 subject to x^2 +xy+y^2 =1.


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Question Number 138851 by bramlexs22 last updated on 19/Apr/21

$F i n d m a x v a l u e o f$ $x^{2} - 3 x y - 2 y^{2} s u b j e c t t o$ $x^{2} + x y + y^{2} = 1 .$
	Answered by ajfour last updated on 19/Apr/21
	$F=x^2 −3xy−2y^2 F=x^2 +3x^2 +3y^2 −3−2y^2 F=4x^2 +y^2 −3 y=−(x/2)±(√((x^2 /4)+1−x^2 )) y=−(x/2)±(√(1−((3x^2 )/4))) F=4x^2 −2−x^2 +(x^2 /2)∓x(√(1−((3x^2 )/4))) F=((7x^2 )/2)−2∓x(√(1−((3x^2 )/4))) (dF/dx)=7x∓{(√(1−((3x^2 )/4)))−((3x^2 )/(4(√(1−((3x^2 )/4)))))}=0 ⇒ 49x^2 =1−((3x^2 )/4)+((9x^4 )/(16−12x^2 ))−((3x^2 )/2) ⇒ ((205x^2 )/4)=1+((9x^4 )/(16−12x^2 )) say x^2 =t ⇒ 205x^2 (4−3x^2 )=16−12x^2 +9x^4 ⇒ 624x^4 −832x^2 +16=0 ⇒ 39x^4 −52x^2 +1=0 x^2 =(2/3)±(√((4/9)−(1/(39)))) x^2 =(2/3)±(7/(3(√(13)))) but x^2 <(4/3) now compare F=((7x^2 )/2)−2+(√(x^2 −((3x^4 )/4))) for x^2 =0, (4/3), (2/3)±(7/(3(√(13)))) F(x^2 =0)=−2 F(x^2 =(4/3))=(8/3)≈2.6667 F(x^2 =(2/3)+(7/(3(√(13)))))≈ 2.737 F(x^2 =(2/3)−(7/(3(√(13)))))≈ −1.793 ⇒ F_(max) =F(x^2 =(2/3)+(7/(3(√(13))))) = (7/2)(((26+7(√(13)))/(39)))+{(((26+7(√(13)))/(39)))[1−(3/4)(((26+7(√(13)))/(39)))]}^(1/2) ≈ 2.737$
	$F = x^{2} - 3 x y - 2 y^{2}$ $F = x^{2} + 3 x^{2} + 3 y^{2} - 3 - 2 y^{2}$ $F = 4 x^{2} + y^{2} - 3$ $y = - \frac{x}{2} \pm \sqrt{\frac{x^{2}}{4} + 1 - x^{2}}$ $y = - \frac{x}{2} \pm \sqrt{1 - \frac{3 x^{2}}{4}}$ $F = 4 x^{2} - 2 - x^{2} + \frac{x^{2}}{2} \mp x \sqrt{1 - \frac{3 x^{2}}{4}}$ $F = \frac{7 x^{2}}{2} - 2 \mp x \sqrt{1 - \frac{3 x^{2}}{4}}$ $\frac{d F}{d x} = 7 x \mp {\sqrt{1 - \frac{3 x^{2}}{4}} - \frac{3 x^{2}}{4 \sqrt{1 - \frac{3 x^{2}}{4}}}} = 0$ $\Rightarrow 49 x^{2} = 1 - \frac{3 x^{2}}{4} + \frac{9 x^{4}}{16 - 12 x^{2}} - \frac{3 x^{2}}{2}$ $\Rightarrow \frac{205 x^{2}}{4} = 1 + \frac{9 x^{4}}{16 - 12 x^{2}}$ $s a y x^{2} = t$ $\Rightarrow 205 x^{2} (4 - 3 x^{2}) = 16 - 12 x^{2} + 9 x^{4}$ $\Rightarrow 624 x^{4} - 832 x^{2} + 16 = 0$ $\Rightarrow 39 x^{4} - 52 x^{2} + 1 = 0$ $x^{2} = \frac{2}{3} \pm \sqrt{\frac{4}{9} - \frac{1}{39}}$ $x^{2} = \frac{2}{3} \pm \frac{7}{3 \sqrt{13}}$ $b u t x^{2} < \frac{4}{3}$ $n o w c o m p a r e$ $F = \frac{7 x^{2}}{2} - 2 + \sqrt{x^{2} - \frac{3 x^{4}}{4}}$ $f o r x^{2} = 0, \frac{4}{3}, \frac{2}{3} \pm \frac{7}{3 \sqrt{13}}$ $F (x^{2} = 0) = - 2$ $F (x^{2} = \frac{4}{3}) = \frac{8}{3} \approx 2.6667$ $F (x^{2} = \frac{2}{3} + \frac{7}{3 \sqrt{13}}) \approx 2.737$ $F (x^{2} = \frac{2}{3} - \frac{7}{3 \sqrt{13}}) \approx - 1.793$ $\Rightarrow F_{m a x} = F (x^{2} = \frac{2}{3} + \frac{7}{3 \sqrt{13}})$ $= \frac{7}{2} (\frac{26 + 7 \sqrt{13}}{39}) + {(\frac{26 + 7 \sqrt{13}}{39}) [1 - \frac{3}{4} (\frac{26 + 7 \sqrt{13}}{39})]}^{1 / 2}$ $\approx 2.737$
	Commented by ajfour last updated on 19/Apr/21

	Answered by mr W last updated on 19/Apr/21

	$s a y k = x^{2} - 3 x y - 2 y^{2}$ $F = x^{2} - 3 x y - 2 y^{2} + λ (x^{2} + x y + y^{2} - 1)$ $\frac{\partial F}{\partial x} = 2 x - 3 y + λ (2 x + y) = 0$ $\Rightarrow 2 (1 + λ) x - (3 - λ) y = 0 . . . (i)$ $\frac{\partial F}{\partial y} = - 3 x - 4 y + λ (x + 2 y) = 0$ $\Rightarrow (λ - 3) x + 2 (λ - 2) y = 0 . . . (i i)$ $\frac{x}{y} = \frac{3 - λ}{2 (1 + λ)} = \frac{2 (λ - 2)}{3 - λ}$ $\Rightarrow 3 λ^{2} + 2 λ - 17 = 0$ $λ = \frac{- 1 \pm 2 \sqrt{13}}{3}$ $\Rightarrow \frac{x}{y} = \frac{2 (λ - 2)}{3 - λ} = \frac{\pm \sqrt{13} - 3}{4} = δ$ $x^{2} + x y + y^{2} = 1 \Rightarrow y^{2} = \frac{1}{δ^{2} + δ + 1}$ $k = (δ^{2} - 3 δ - 2) y^{2} = \frac{δ^{2} - 3 δ - 2}{δ^{2} + δ + 1}$ $w i t h δ = \frac{\sqrt{13} - 3}{4} :$ $k_{m i n} = \frac{1 - 2 \sqrt{13}}{3} \approx - 2.070368$ $w i t h δ = \frac{- \sqrt{13} - 3}{4} :$ $k_{m a x} = \frac{1 + 2 \sqrt{13}}{3} \approx 2.737034$
	Commented by mr W last updated on 19/Apr/21

	Commented by mr W last updated on 19/Apr/21

	$g r e e n : x^{2} + x y + y^{2} = 1$ $r e d / p i n k : x^{2} - 3 x y - 2 y^{2} = k$
	Answered by ajfour last updated on 19/Apr/21

	$l e t y = t x$ $x^{2} (1 + t + t^{2}) = 1$ $f (x, t) = f = x^{2} (1 - 3 t - 2 t^{2})$ $f = \frac{1 - 3 t - 2 t^{2}}{t^{2} + t + 1}$ $f = - 2 - \frac{t - 3}{t^{2} + t + 1}$ $f = - 2 - \frac{1}{t + 4 + \frac{13}{t - 3}}$ $f = - 2 - \frac{1}{t - 3 + \frac{13}{t - 3} + 7}$ $f i s f_{m a x} i f h (t) = t - 3 + \frac{13}{t - 3} + 7$ $i s n e g a t i v e w i t h m i n i m u m$ $a b s o l u t e v a l u e .$ $h^{'} (t) = 1 - \frac{13}{{(t - 3)}^{2}} = 0$ $a t t - 3 = - \sqrt{13}$ $h_{m i n} = 7 - 2 \sqrt{13}$ $f_{m a x} = - 2 + \frac{1}{2 \sqrt{13} - 7}$ $= - 2 + \frac{2 \sqrt{13} + 7}{3}$ $f_{m a x} = \frac{2 \sqrt{13} + 1}{3} \approx 2.73703418$
	Commented by ajfour last updated on 19/Apr/21

	Commented by mr W last updated on 20/Apr/21

	$n i c e a p p r o a c h!$
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