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

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\textbf{(E)}\ 9 </math>
 
\textbf{(E)}\ 9 </math>
  
== Solution ==
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== 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 00:53, 10 November 2015

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 equations could be deduced: $(x+y)+(x+y)$

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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