$x+y=10⇒x=10-y$ Khi đó $C=4x^2+9y^2$ $=4(10-y)^2+9y^2$ $=400-80y+4y^2+9y^2$ $=13y^2-80y+400$ $=13.(y^2-\dfrac{80}{13}y+\dfrac{400}{13})$ $=13.(y^2-2.\dfrac{40}{13}.y+\dfrac{1600}{169}+\dfrac{3600}{169})$ $=13.(y-\dfrac{40}{3})^2+\dfrac{3600}{13}≥\dfrac{3600}{13}$ Dấu $=$ xảy ra $⇔y-\dfrac{40}{3}=0⇔y=\dfrac{40}{3};x=\dfrac{-10}{3}$ Bình luận
$x+y=10⇒x=10-y$
Khi đó $C=4x^2+9y^2$
$=4(10-y)^2+9y^2$
$=400-80y+4y^2+9y^2$
$=13y^2-80y+400$
$=13.(y^2-\dfrac{80}{13}y+\dfrac{400}{13})$
$=13.(y^2-2.\dfrac{40}{13}.y+\dfrac{1600}{169}+\dfrac{3600}{169})$
$=13.(y-\dfrac{40}{3})^2+\dfrac{3600}{13}≥\dfrac{3600}{13}$
Dấu $=$ xảy ra $⇔y-\dfrac{40}{3}=0⇔y=\dfrac{40}{3};x=\dfrac{-10}{3}$