AMC12 2011 A

AMC12 2011 A · Q18

AMC12 2011 A · Q18. It mainly tests Absolute value, Inequalities (AM-GM etc. basic).

Suppose that $\left|x+y\right|+\left|x-y\right|=2$. What is the maximum possible value of $x^2-6x+y^2$?

设 $\left|x+y\right|+\left|x-y\right|=2$。$x^2-6x+y^2$ 的最大可能值是多少？

(A) 5 5

(B) 6 6

(D) 8 8

(E) 9 9

Answer

Correct choice: (D)

正确答案：（D）

Solution

Plugging in some values, we see that the graph of the equation $|x+y|+|x-y| = 2$ is a square bounded by $x= \pm 1$ and $y = \pm 1$. Notice that $x^2 - 6x + y^2 = (x-3)^2 + y^2 - 9$ means the square of the distance from a point $(x,y)$ to point $(3,0)$ minus 9. To maximize that value, we need to choose the point in the feasible region farthest from point $(3,0)$, which is $(-1, \pm 1)$. Either one, when substituting into the function, yields $\boxed{\textbf{(D)}\ 8}$.

代入一些数值可知，方程 $|x+y|+|x-y| = 2$ 的图像是由 $x= \pm 1$ 与 $y = \pm 1$ 围成的正方形。注意到 $x^2 - 6x + y^2 = (x-3)^2 + y^2 - 9$，这表示点 $(x,y)$ 到点 $(3,0)$ 的距离平方再减去 9。要使该值最大，需要在可行区域内选取离点 $(3,0)$ 最远的点，即 $(-1, \pm 1)$。将其代入原式可得 $\boxed{\textbf{(D)}\ 8}$。

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