2011 AMC 12A Problem 18

Attempt Problem 18 of the 2011 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 12A solutions, or check the answer key.

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18.

Suppose that

|x + y| + |x - y| = 2.

What is the maximum possible value of

x^2 - 6x + y^2?

5

6

7

8

9

The equation

|x+y| + |x-y| = 2

means

\max(|x|, |y|) = 1,

the boundary of the square

[-1,1]^2

On that square,

x^2 - 6x + y^2

is largest when

x = -1

and

y = \pm 1

The identity

|x+y| + |x-y|

= 2\max(|x|, |y|)

turns the condition into

\max(|x|, |y|) = 1,

the boundary of the square with

|x| \le 1

and

|y| \le 1.

On this region

x^2 - 6x + y^2

increases as

x

decreases and as

y^2

increases, so the maximum is at

x = -1,

y = \pm 1:

1 + 6 + 1 = 8.

Thus, the correct answer is D.

Problem 18 in other years