F(x,y)=x^2 −2xy+6y^2 −12x+2y+45 find x &y such that F(x,y) minimum


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Question Number 157247 by john_santu last updated on 21/Oct/21

$F (x, y) = x^{2} - 2 x y + 6 y^{2} - 12 x + 2 y + 45$ $f i n d x & y s u c h t h a t F (x, y) m i n i m u m$
	Answered by FongXD last updated on 21/Oct/21

	$Given : f (x, y) = x^{2} - 2 xy + {6 y}^{2} - 12 x + 2 y + 45$ $\Leftrightarrow f (x, y) = (x^{2} + y^{2} + 36 - 2 xy - 12 x + 12 y) + ({5 y}^{2} - 10 y + 9)$ $\Leftrightarrow f (x, y) = {(x - y - 6)}^{2} + 5 {(y - 1)}^{2} + 4$ $since {(x - y - 6)}^{2} + 5 {(y - 1)}^{2} ⩾ 0, \forall x, y \in R$ $therefore, Min [f (x, y)] = 4, when y = 1 and x = 7$ $so . \begin{matrix} Min [f (x, y)] = 4 which occurs when x = 7 and y = 1 \end{matrix}$
	Answered by mr W last updated on 21/Oct/21

	$x^{2} - 2 x y + 6 y^{2} - 12 x + 2 y + 45 = k$ $x^{2} - 2 (y + 6) x + 6 y^{2} + 2 y + 45 - k = 0$ ${(y + 6)}^{2} - (6 y^{2} + 2 y + 45 - k) ⩾ 0$ $5 y^{2} - 10 y + 9 - k ⩽ 0$ $10^{2} - 4 \times 5 (9 - k) ⩾ 0$ $k ⩾ 4$ $\Rightarrow F {(x, y)}_{m i n} = 4$ $y^{2} - 2 y + 1 = 0 \Rightarrow y = 1$ $x^{2} - 14 x + 49 = 0 \Rightarrow x = 7$
	Answered by qaz last updated on 21/Oct/21
	${ (((∂F/∂x)=2x−2y−12=0)),(((∂F/∂y)=−2x+12y+2=0)) :} ⇒M=(x,y)=(7,1) A=(∂^2 F/∂x^2 )∣_M =2 B=(∂^2 F/∂y^2 )∣_M =12 C=(∂^2 F/(∂x∂y))∣_M =−2 ∵ A>0 AB−C^2 >0 ∴ MinF(x,y)=F(7,1)=4$
	${\begin{cases} \frac{\partial F}{\partial x} = 2 x - 2 y - 12 = 0 \\ \frac{\partial F}{\partial y} = - 2 x + 12 y + 2 = 0 \end{cases} \Rightarrow M = (x, y) = (7, 1)$ $A = \frac{\partial^{2} F}{\partial x^{2}} ∣_{M} = 2 B = \frac{\partial^{2} F}{\partial y^{2}} ∣_{M} = 12 C = \frac{\partial^{2} F}{\partial x \partial y} ∣_{M} = - 2$ $∵ A > 0 AB - C^{2} > 0$ $∴ MinF (x, y) = F (7, 1) = 4$
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