Difference between revisions of "2011 AMC 12A Problems/Problem 18"
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\textbf{(E)}\ 9 </math> | \textbf{(E)}\ 9 </math> | ||
| − | == Solution == | + | == Solution 1 == |
Plugging in some values, we see that the graph of the equation <math>|x+y|+|x-y| = 2</math> is a square bounded by <math>x= \pm 1</math> and <math>y = \pm 1</math>. | Plugging in some values, we see that the graph of the equation <math>|x+y|+|x-y| = 2</math> is a square bounded by <math>x= \pm 1</math> and <math>y = \pm 1</math>. | ||
Notice that <math>x^2 - 6x + y^2 = (x-3)^2 + y^2 - 9</math> means the square of the distance from a point <math>(x,y)</math> to point <math>(3,0)</math> minus 9. To maximize that value, we need to choose the point in the feasible region farthest from point <math>(3,0)</math>, which is <math>(-1, \pm 1)</math>. Either one, when substituting into the function, yields <math>8 \rightarrow \boxed{(D)}</math>. | Notice that <math>x^2 - 6x + y^2 = (x-3)^2 + y^2 - 9</math> means the square of the distance from a point <math>(x,y)</math> to point <math>(3,0)</math> minus 9. To maximize that value, we need to choose the point in the feasible region farthest from point <math>(3,0)</math>, which is <math>(-1, \pm 1)</math>. Either one, when substituting into the function, yields <math>8 \rightarrow \boxed{(D)}</math>. | ||
| + | |||
| + | == 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)</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 23:53, 9 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:
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
