find-max-and-min-value-of-f-x-y-4x-2-8xy-9y-2-8x-24y-4-

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) = 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 \dots \dots (1)

f_{y} = \frac{d}{d y} f

f_{y} = 8 x + 18 y - 24

f_{y} = 0

8 x + 18 y - 24 = 0 \dots \dots (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})

Facebook Tweet Pin