Difference between revisions of "2011 AMC 12A Problems/Problem 18"
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== Solution 2 == | == Solution 2 == | ||
− | Since the equation <math>|x+y|+|x-y| = 2</math> is dealing with absolute values, the following equations could be deduced: <math>(x+y)+(x+y)=2</math>,<math>(x+y)-(x-y)=2</math>, <math>-(x+y)+(x-y)=2</math>, and <math>-(x+y)-(x-y)=2</math>. Simplifying would give <math>x=1</math>, <math>y=1</math>, <math>y=-1</math>, and <math>x=-1</math>. In <math>x^2-6x+y^2</math>, it does not matter whether x or y is <math>-1</math> or <math>1</math>. To maximize <math>-6x</math>, though, <math>x</math> would have to be -1. Therefore, when <math>x=-1</math> and <math>y=-1</math> or <math>y=1</math>, the equation evaluates to <math>\boxed{(D)}8</math> | + | Since the equation <math>|x+y|+|x-y| = 2</math> is dealing with absolute values, the following equations could be deduced: <math>(x+y)+(x+y)=2</math>,<math>(x+y)-(x-y)=2</math>, <math>-(x+y)+(x-y)=2</math>, and <math>-(x+y)-(x-y)=2</math>. Simplifying would give <math>x=1</math>, <math>y=1</math>, <math>y=-1</math>, and <math>x=-1</math>. In <math>x^2-6x+y^2</math>, it does not matter whether x or y is <math>-1</math> or <math>1</math>. To maximize <math>-6x</math>, though, <math>x</math> would have to be -1. Therefore, when <math>x=-1</math> and <math>y=-1</math> or <math>y=1</math>, the equation evaluates to <math>\boxed{(D)} 8</math> |
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=17|num-a=19|ab=A}} | {{AMC12 box|year=2011|num-b=17|num-a=19|ab=A}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:04, 10 November 2015
Contents
Problem
Suppose that . What is the maximum possible value of
?
Solution 1
Plugging in some values, we see that the graph of the equation is a square bounded by
and
.
Notice that means the square of the distance from a point
to point
minus 9. To maximize that value, we need to choose the point in the feasible region farthest from point
, which is
. Either one, when substituting into the function, yields
.
Solution 2
Since the equation is dealing with absolute values, the following equations could be deduced:
,
,
, and
. Simplifying would give
,
,
, and
. In
, it does not matter whether x or y is
or
. To maximize
, though,
would have to be -1. Therefore, when
and
or
, the equation evaluates to
See also
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 |
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