b2 - 5.20 (optimering)

Återigen har lite svårt med optimeringsproblemen, svaret enligt facit är f_min=-16^(2)/13 och f_max=4.

Här är min lösning hittils:

$\begin{array}{l} L ö s : f (x, y) = 4 x^{2} y^{2} - 2 x y^{4} - 3 x^{2}, 0 \leq x \leq 2, 0 \leq y \leq 2 \\ 1 . S t a t i o n ä r a p u n k t e r : \\ f^{'} x = 8 x y^{2} - 2 y^{4} - 6 x = 2 (4 x y^{2} - 2 y^{4} - 3 x) \\ f^{'} y = 8 x^{2} y - 8 x y^{3} = 8 x y (x - y^{2}) = 0, x y (x - y^{2}) = 0 \\ S t a t i o n ä r p u n k t : (x, y) = (0, 0), [x = y^{2}] \\ f (0, 0) = [0] (i n t r e s s a n t p u n k t) \\ 8 y^{4} - 2 y^{4} - 6 y^{2} = 6 y^{4} - 6 y^{2} = 6 y^{2} (y^{2} - 1), [y = 1], [x = 1] \\ S t a t i o n ä r p u n k t : (x, y) = (1, 1) \\ f (1, 1) = [- 1] (i n t r e s s a n t p u n k t) \\ 2 . R a n d e n : \\ g_{1} (x) = f (x, 0) = - 3 x^{2}, g_{1} (x) = - 3 x^{2} = 0, [x = 0] \\ f (0, 0) (i n t r e s s a n t p u n k t, r e d a n b e r ä k n a d) \\ g_{2} (x) = f (x, 2) = 13 x^{2} - 32 x, g_{2} (x) = x (13 x - 32) = 0, [x_{1} = 0] [x_{2} = \frac{32}{13}] \\ f (0, 2) = [0] (i n t r e s s a n t p u n k t) \\ f (\frac{32}{13}, 2) (e j i n t r e s s a n t d å \frac{32}{13} > 2) \\ f (0, y) = [0] (a l l a ä r i n t r e s s a n t a p u n k t e r) \\ f (2, y) = 16 y^{2} - 4 y^{4} - 12 = 0, 4 y^{2} - y^{4} - 3 = 0, [r = y^{2}] \\ P Q - f o r m e l : - r^{2} + 4 r - 3 = 0, r^{2} - 4 r + 3 = 0 \\ r = 2 + - {(4 - 3)}^{\frac{1}{2}} = 2 + - 1, [r_{1} = 3] [r_{2} = 1], [y_{1} = 1], [y_{2} = - 1], [y_{3} = 3^{\frac{1}{2}}] \\ f (2, 1) = [0] (i n t r e s s a n t) \\ f (2, - 1) (e j i n t r e s s a n t d å - 1 < 0) \\ f (2, 3^{\frac{1}{2}}) = [0] (i n t r e s s a n t) \\ H ö r n : \\ f (0, 0) (i n t r e s s a n t p u n k t, r e d a n u t r ä k n a t) \\ f (0, 2) (i n t r e s s a n t p u n k t, r e d a n u t r ä k n a t) \\ f (2, 2) = [- 12] (i n t r e s s a n t p u n k t) \\ f (2, 0) = [- 12] (i n t r e s s a n t p u n k t) \\ A l l a i n t r e s s a n t a p u n k t e r : \\ f (0, 0) = [0], f (1, 1) = [- 1], f (0, 2) = [0], f (0, y) = [0], \\ f (2, 1) = [0], f (2, 3^{\frac{1}{2}}) = [0], f (2, 2) = [- 12], f (2, 0) = [- 12] \\ S v a r : f_{\min} = - 12, f_{\max} = 0 \end{array}$

Borde du inte undersöka då x= y²?

Edit: Jag missade att du visst gjorde det. Och att du råkade göra ett slarvfel.

$\begin{array}{l} S t a t i o n ä r p u n k t : (x, y) = (0, 0), [x = y^{2}] \\ f (0, 0) = [0] (i n t r e s s a n t p u n k t) \\ 8 y^{4} - 4 y^{4} - 6 y^{2} = 4 y^{4} - 6 y^{2} = 2 y^{2} (2 y^{2} - 3), [y = \pm \sqrt{\frac{3}{2}}], [x = 1] \end{array}$

Tillägg: 20 apr 2024 07:52

[x=1] stämmer inte heller nu.

Svara Avbryt