find max and min value of f(x,y) = 4x^2 +8xy+9y^2 −8x−24y+4


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Question Number 119240 by benjo_mathlover last updated on 23/Oct/20

$f i n d m a x a n d m i n v a l u e o f$ $f (x, y) = 4 x^{2} + 8 x y + 9 y^{2} - 8 x - 24 y + 4$
	Answered by 1549442205PVT last updated on 23/Oct/20
	$f(x,y) = 4x^2 +8xy+9y^2 −8x−24y+4 =[4x^2 +2.2x(2y−2)+4y^2 −8y+4]+5y^2 −16y =(2x+2y−2)^2 +5y^2 −16y =4(x+y−1)^2 +5(y−(8/5))^2 −((64)/5)≥((−64)/5) ⇒f_(min) (x,y)=((−64)/5) when { ((y−8/5=0)),((x+y−1=0)) :}⇔ { ((x=−3/5)),((y=8/5)) :} f_(max) (x,y)=+∞$
	$f (x, y) = 4 x^{2} + 8 x y + 9 y^{2} - 8 x - 24 y + 4$ $= [{4 x}^{2} + 2 . 2 x (2 y - 2) + {4 y}^{2} - 8 y + 4] + {5 y}^{2} - 16 y$ $= {(2 x + 2 y - 2)}^{2} + {5 y}^{2} - 16 y$ $= 4 {(x + y - 1)}^{2} + 5 {(y - \frac{8}{5})}^{2} - \frac{64}{5} ⩾ \frac{- 64}{5}$ $\Rightarrow f_{\min} (x, y) = \frac{- 64}{5} when$ ${\begin{cases} y - 8 / 5 = 0 \\ x + y - 1 = 0 \end{cases} \Leftrightarrow {\begin{cases} x = - 3 / 5 \\ y = 8 / 5 \end{cases}$ $f_{\max} (x, y) = + \infty$
	Answered by ebi last updated on 23/Oct/20

	$f_{x} = \frac{d}{d x} f$ $f_{x} = 8 x + 8 y - 8$ $f_{x} = 0$ $8 x + 8 y - 8 = 0 . . . . . . (1)$ $f_{y} = \frac{d}{d y} f$ $f_{y} = 8 x + 18 y - 24$ $f_{y} = 0$ $8 x + 18 y - 24 = 0 . . . . . . (2)$ $(1) - (2) : - 10 y = - 16 \to y = \frac{8}{5}$ $∴ 8 x + 8 (\frac{8}{5}) = 8 \to x = - \frac{3}{5}$ $(- \frac{3}{5}, \frac{8}{5}) w h e n f_{x} = 0 a n d f_{y} = 0$ $D - T e s t$ $D = f_{x x} \cdot f_{y y} - {(f_{x y})}^{2}$ $f_{x x} = 8$ $f_{y y} = 18$ $f_{x y} = 8$ $D = 8 (18) - 8^{2} = 80$ $s i n c e D = 80 > 0 a n d f_{x x} = 8 > 0,$ $t h e n f h a s a m i n i m u m a t (- \frac{3}{5}, \frac{8}{5})$
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