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Question Number 161626 by ZiYangLee last updated on 20/Dec/21

Commented by TheSupreme last updated on 20/Dec/21
$f(x,y,λ)=x−2y+λ(4x^2 +9y^2 −16x−54y+61) { ((1+λ(8x−16)=0)),((−2+λ(18y−54)=0)),((4x^2 +9y^2 −16x−54y+61=0)) :} x=2−(1/(8λ)) y=3+(1/(9λ)) (4−(1/λ))^2 +(9+(1/λ))^2 −(32−(2/λ))−(162+(6/λ))+61=0 −35+(2/λ^2 )+(2/λ)=0 λ=(1/(−2±(√(144))))=(1/(2±12))=(1/(14)),−(1/(10)) x−2y=−4−(2/(9λ))−(1/(8λ))=−4−((25)/(72λ)) max = −4+((25)/(72))14=((−288+350)/(72))=1$
$f (x, y, λ) = x - 2 y + λ (4 x^{2} + 9 y^{2} - 16 x - 54 y + 61)$ ${\begin{cases} 1 + λ (8 x - 16) = 0 \\ - 2 + λ (18 y - 54) = 0 \\ 4 x^{2} + 9 y^{2} - 16 x - 54 y + 61 = 0 \end{cases}$ $x = 2 - \frac{1}{8 λ}$ $y = 3 + \frac{1}{9 λ}$ ${(4 - \frac{1}{λ})}^{2} + {(9 + \frac{1}{λ})}^{2} - (32 - \frac{2}{λ}) - (162 + \frac{6}{λ}) + 61 = 0$ $- 35 + \frac{2}{λ^{2}} + \frac{2}{λ} = 0$ $λ = \frac{1}{- 2 \pm \sqrt{144}} = \frac{1}{2 \pm 12} = \frac{1}{14}, - \frac{1}{10}$ $x - 2 y = - 4 - \frac{2}{9 λ} - \frac{1}{8 λ} = - 4 - \frac{25}{72 λ}$ $m a x = - 4 + \frac{25}{72} 14 = \frac{- 288 + 350}{72} = 1$
	Answered by mr W last updated on 20/Dec/21

	$s a y x - 2 y = k$ $\Rightarrow x = k + 2 y$ $4 {(k + 2 y)}^{2} + 9 y^{2} - 16 (k + 2 y) - 54 y + 61 = 0$ $25 y^{2} - 2 (43 - 8 k) y + 4 k^{2} - 16 k + 61 = 0$ $Δ = {(43 - 8 k)}^{2} - 25 (4 k^{2} - 16 k + 61) ⩾ 0$ $k^{2} + 8 k - 9 ⩽ 0$ $(k + 9) (k - 1) ⩽ 0$ $- 9 ⩽ k ⩽ 1$ $\Rightarrow {(x - 2 y)}_{m i n} = k_{m i n} = - 9$ $\Rightarrow {(x - 2 y)}_{m a x} = k_{m a x} = 1$
	Commented by mr W last updated on 20/Dec/21

	$x - 2 y = - 9 a n d x - 2 y = 1 a r e t w o$ $l i n e s w h i c h t a n g e n t t h e e l l i p s e .$
	Commented by mr W last updated on 20/Dec/21

	Commented by ZiYangLee last updated on 20/Dec/21

	$Thank you!$
	Answered by aleks041103 last updated on 20/Dec/21

	$4 x^{2} + 9 y^{2} - 16 x - 54 y + 61 =$ $= {(2 x)}^{2} - 2 (2 x) (4) + 4^{2} + {(3 y)}^{2} - 2 (3 y) (9) + 9^{2} + 61 - 4^{2} - 9^{2} =$ $= {(2 x - 4)}^{2} + {(3 y - 9)}^{2} + 61 - 16 - 81 = 0$ $\Rightarrow {(2 x - 4)}^{2} + {(3 y - 9)}^{2} = 6^{2}$ $\Rightarrow {(\frac{x - 2}{3})}^{2} + {(\frac{y - 3}{2})}^{2} = 1$ $\Rightarrow \frac{x - 2}{3} = c o s t$ $\frac{y - 3}{2} = s i n t$ $\Rightarrow x = 2 + 3 c o s t$ $y = 3 + 2 s i n t$ $\Rightarrow x - 2 y = 3 c o s t - 4 s i n t - 4 =$ $= 5 (\frac{3}{5} c o s t - \frac{4}{5} s i n t) - 4 =$ $= 5 (c o s θ c o s t - s i n θ s i n t) - 4 =$ $= 5 c o s (θ + t) - 4$ $\Rightarrow m a x (x - 2 y) = 5 m a x (c o s (θ + t)) - 4 =$ $= 5 - 4 = 1$
	Commented by peter frank last updated on 20/Dec/21

	$thank you$
	Answered by mr W last updated on 20/Dec/21

	$A N O T H E R M E T H O D$ $4 (x^{2} - 4 x + 4) + 9 (y^{2} - 6 y + 9) = 36$ $\frac{{(x - 2)}^{2}}{3^{2}} + \frac{{(y - 3)}^{2}}{2^{2}} = 1$ $c e n t e r o f e l l i p s e i s (2, 3)$ $s e m i a x i s e s : a = 3, b = 2$ $s a y x - 2 y = k$ $l i n e x - 2 y - k = 0 s h o u l d t a n g e n t t h e$ $e l l i p s e .$ $3^{2} \times 1^{2} + 2^{2} \times {(- 2)}^{2} = {(2 \times 1 - 2 \times 3 - k)}^{2}$ $25 = {(4 + k)}^{2}$ $k + 4 = \pm 5$ $\Rightarrow k = - 4 \pm 5 = - 9 o r 1$
	Commented by peter frank last updated on 20/Dec/21

	$thank you$
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