Find minimum value of 2x^2 +2xy+4y+5y^2 −x for x and y real numbers


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Question Number 188806 by horsebrand11 last updated on 07/Mar/23

$F i n d m i n i m u m v a l u e o f$ $2 x^{2} + 2 x y + 4 y + 5 y^{2} - x$ $f o r x a n d y r e a l n u m b e r s$
	Answered by mr W last updated on 07/Mar/23

	$M e t h o d I$ $k = 2 x^{2} + 2 x y + 4 y + 5 y^{2} - x$ $2 x^{2} + (2 y - 1) x + 5 y^{2} + 4 y - k = 0$ $Δ = {(2 y - 1)}^{2} - 8 (5 y^{2} + 4 y - k) ⩾ 0$ $- 36 y^{2} - 36 y + 8 k + 1 ⩾ 0$ $36 y^{2} + 36 y - 8 k - 1 ⩽ 0$ $Δ = 36^{2} + 4 \times 36 (8 k + 1) ⩾ 0$ $\Rightarrow k ⩾ - \frac{5}{4}$ ${(2 x^{2} + 2 x y + 4 y + 5 y^{2} - x)}_{m i n} = - \frac{5}{4}$
	Answered by mr W last updated on 07/Mar/23

	$M e t h o d I I$ $k = 2 x^{2} + 2 x y + 4 y + 5 y^{2} - x$ $\frac{\partial k}{\partial x} = 4 x + 2 y - 1 = 0 . . . (i)$ $\frac{\partial k}{\partial y} = 2 x + 4 + 10 y = 0 . . . (i i)$ $\Rightarrow x = \frac{1}{2}, y = - \frac{1}{2}$ $k_{m i n} = 2 \times \frac{1}{4} - \frac{1}{2} - 2 + \frac{5}{4} - \frac{1}{2} = - \frac{5}{4} ✓$ $({i t}^{'} s m i n i m u m, b e c a u s e \frac{\partial^{2} k}{\partial x^{2}} > 0, \frac{\partial^{2} k}{\partial y^{2}} > 0)$
	Commented by ARUNG_Brandon_MBU last updated on 07/Mar/23
	Sir can you check my reply on your comment in Q188362
	Commented by mr W last updated on 07/Mar/23

	$y e s, y o u g o t t h e r i g h t r e s u l t! t h a n k s!$ $s o r r y, i {d o n}^{'} t g e t n o t i f i e d a b o u t y o u r$ $r e p l y .$
	Commented by ARUNG_Brandon_MBU last updated on 07/Mar/23

	$OK Sir .$ $Me too . I {don}^{'} t get notified about replies .$
	Commented by mr W last updated on 08/Mar/23

	$c a n y o u c h e c k m y a n s w e r i n Q 111118$ $w i t h a p r o g r a m ? t h a n k s s i r!$
	Commented by ARUNG_Brandon_MBU last updated on 09/Mar/23
	The system says "Question deleted from forum."
	Commented by mr W last updated on 09/Mar/23

	$You can't use 'macro parameter character #' in math mode$ $i m i s t o o k t h e I D o f t h a t u s e r f o r t h e$ $q u e s t i o n n u m b e r .$
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