Cho $x,y$ là 2 số thỏa mãn đồng thời $x\geq0$,$y\geq0$,$2x+3y\leq 6$ và $2x+y\leq4$.
Tìm min và max của $K=x^2-2x-y$
$K=x^2-2x-y=x^2-2x+1-y-1=(x-1)^2-y-1$
Có: $2x+3y\leq6$ và $2x+y\leq4$, nên $2x+3y-2x-y=2y\leq2$ nên $y\leq1$
Nên, $K=(x-1)^2-y-1 \geq 0 - 2 -1 = -3$
Dấu bằng xảy ra khi và chỉ khi $y=1$ và $x=1$
$K=x(x-2)-y$
Xét: Vì $2x+y\leq4$ nên $2x\leq4$(Vì $y\geq0$)
Hay $x\leq2$
Nên, $0\leq x\leq2$
Nên, $x-2\leq0$
Mà $x\geq0$
Nên $x(x-2)\leq0$
Vậy, $K=x(x-2)-y\leq0-0=0$
Dấu bằng xảy ra khi và chỉ khi $x=0$ và $y=0$ hay $x=2$ và $y=0$

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