2011 AMC 12A Problems/Problem 18

Problem

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

$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$

Solution 1

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 $8 \rightarrow \boxed{(D)}$ .

Solution 2

Since the equation $|x+y|+|x-y| = 2$ is dealing with absolute values, the following could be deduced: $(x+y)+(x-y)=2$ , $(x+y)-(x-y)=2$ , $-(x+y)+(x-y)=2$ , and $-(x+y)-(x-y)=2$ . Simplifying would give $x=1$ , $y=1$ , $y=-1$ , and $x=-1$ . In $x^2-6x+y^2$ , we care most about $-6x,$ since both $x^2$ and $y^2$ are non-negative. To maximize $-6x$ , though, $x$ would have to be -1. Therefore, when $x=-1$ and $y=-1$ or $y=1$ , the equation evaluates to $\boxed{(D)}$ $8$ .

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AMC 12 Problems and Solutions

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2011 AMC 12A Problems/Problem 18

Contents

Problem

Solution 1

Solution 2

See also