Find-minimum-value-of-2x-2-2xy-4y-5y-2-x-for-x-and-y-real-numbers-

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 \dots (i)

\frac{\partial k}{\partial y} = 2 x + 4 + 10 y = 0 \dots (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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