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

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AMC 12 Problems and Solutions

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