Tìm min của `K=2x^2+9y^2-6xy-6x-12y+2004`

Đáp án:

$\min K = 1975 \Leftrightarrow (x;y) = \left(5;\dfrac{7}{3}\right)$

Giải thích các bước giải:

$\begin{array}{l}K = 2x^2 + 9y^2- 6xy – 6x – 12y + 2004\\ = \dfrac{1}{2}(4x^2 + 9y^2 +9 – 12xy + 18y – 12x)+ \dfrac{9}{2}\left(y^2 – \dfrac{14}{3}y + \dfrac{49}{9}\right) + 1975\\ = \dfrac{1}{2}(2x – 3y – 3)^2 + \dfrac{9}{2}\left(y – \dfrac{7}{3}\right)^2 + 1975\\ Ta\,\,có:\\ \begin{cases}(2x – 3y – 3)^2 \geq 0, \,\forall x,y\\\left(y – \dfrac{7}{3}\right)^2 \geq 0, \,\forall y\end{cases}\\ Do\,\,đó:\\ \dfrac{1}{2}(2x – 3y – 3)^2 + \dfrac{9}{2}\left(y – \dfrac{7}{3}\right)^2 + 1975 \geq 1975\\ \text{Dấu = xảy ra}\, \Leftrightarrow \begin{cases}2x – 3y – 3 = 0\\y – \dfrac{7}{3} = 0\end{cases} \Leftrightarrow \begin{cases}x = 5\\y = \dfrac{7}{3}\end{cases}\\ Vậy\,\,\min K = 1975 \Leftrightarrow (x;y) = \left(5;\dfrac{7}{3}\right) \end{array}$