Difference between revisions of "2011 AMC 12A Problems/Problem 18"

Revision as of 20:52, 3 July 2013

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

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

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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 == See also ==
 {{AMC12 box|year=2011|num-b=17|num-a=19|ab=A}}
+{{MAA Notice}}

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Difference between revisions of "2011 AMC 12A Problems/Problem 18"

Revision as of 20:52, 3 July 2013

Problem

Solution

See also