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Question Number 137650 by bemath last updated on 05/Apr/21

	Answered by bemath last updated on 05/Apr/21
	$By Langrange multiplier f(x,y,λ)= x^2 +y^2 +λ(6x^2 +2xy+6y^2 −9) (∂f/∂x) = 2x+λ(12x+2y)=0 ⇒x+λ(6x+y)=0⇒λ=−(x/(6x+y)) (∂f/∂y)=2y+λ(2x+12y)=0 ⇒y+λ(x+6y)=0⇒λ=−(y/(x+6y)) ⇔ (x/(6x+y)) = (y/(x+6y)) ⇒x^2 +6xy=y^2 +6xy it follows that x^2 =y^2 or { ((x=y)),((x=−y)) :} (∂f/∂λ)= 6x^2 +2xy+6y^2 =9 ⇒12x^2 +2xy=9 case(1)x=y ⇒14x^2 =9 f(x,y) = x^2 +y^2 = (9/7) case(2)x=−y ⇒10x^2 =9 f(x,y) = (9/5) Thus maximum value is (9/5)$
	$B y L a n g r a n g e m u l t i p l i e r$ $f (x, y, λ) = x^{2} + y^{2} + λ (6 x^{2} + 2 x y + 6 y^{2} - 9)$ $\frac{\partial f}{\partial x} = 2 x + λ (12 x + 2 y) = 0$ $\Rightarrow x + λ (6 x + y) = 0 \Rightarrow λ = - \frac{x}{6 x + y}$ $\frac{\partial f}{\partial y} = 2 y + λ (2 x + 12 y) = 0$ $\Rightarrow y + λ (x + 6 y) = 0 \Rightarrow λ = - \frac{y}{x + 6 y}$ $\Leftrightarrow \frac{x}{6 x + y} = \frac{y}{x + 6 y}$ $\Rightarrow x^{2} + 6 x y = y^{2} + 6 x y$ $i t f o l l o w s t h a t x^{2} = y^{2} o r {\begin{cases} x = y \\ x = - y \end{cases}$ $\frac{\partial f}{\partial λ} = 6 x^{2} + 2 x y + 6 y^{2} = 9$ $\Rightarrow 12 x^{2} + 2 x y = 9$ $c a s e (1) x = y \Rightarrow 14 x^{2} = 9$ $f (x, y) = x^{2} + y^{2} = \frac{9}{7}$ $c a s e (2) x = - y \Rightarrow 10 x^{2} = 9$ $f (x, y) = \frac{9}{5}$ $T h u s m a x i m u m v a l u e i s \frac{9}{5}$
	Commented by bemath last updated on 05/Apr/21

	Answered by mr W last updated on 05/Apr/21

	$f o r a, b > 0 w e h a v e$ $\frac{a + b}{2} ⩾ \sqrt{a b}$ $9 = 6 (x^{2} + y^{2}) + 2 x y ⩾ 6 (x^{2} + y^{2}) - (x^{2} + y^{2}) = 5 (x^{2} + y^{2})$ $\Rightarrow x^{2} + y^{2} ⩽ \frac{9}{5}$ $\Rightarrow m a x i m u m o f x^{2} + y^{2} = \frac{9}{5}$ $9 = 6 (x^{2} + y^{2}) + 2 x y ⩽ 6 (x^{2} + y^{2}) + (x^{2} + y^{2}) = 7 (x^{2} + y^{2})$ $\Rightarrow x^{2} + y^{2} ⩾ \frac{9}{7}$ $\Rightarrow m i n i m u m o f x^{2} + y^{2} = \frac{9}{7}$
	Commented by bemath last updated on 05/Apr/21

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