Vì \(x+4y=5\Rightarrow x=5-4y\)
\(M=4x^2+y^2=4\left(5-4y\right)^2+y^2=65y^2-160y+100\)
\(=65\left(y-\dfrac{16}{13}\right)^2+\dfrac{20}{13}\ge\dfrac{20}{13}\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{13}\\y=\dfrac{16}{13}\end{matrix}\right.\)
\(x+4y=5\Leftrightarrow x=5-4y\)
Khi đó :
\(M=4\left(5-4y\right)^2+y^2=4\left(25-40y+16y^2\right)+y^2=65y^2-160+100=65\left(y^2-2.\dfrac{16}{13}+\dfrac{256}{169}\right)+\dfrac{20}{13}=65\left(y-\dfrac{16}{13}\right)^2+\dfrac{20}{13}\ge\dfrac{20}{13}\)
Dấu ''='' xảy ra khi \(y-\dfrac{16}{13}=0\Leftrightarrow y=\dfrac{16}{13}\) --> x = 5 - 4. 16/13 = 1/13
Vậy min của M là 20/13 khi x = 1/13 và y = 16/13
