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

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== Problem ==
 
== Problem ==
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Suppose that <math>\left|x+y\right|+\left|x-y\right|=2</math>. What is the maximum possible value of <math>x^2-6x+y^2</math>?
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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Suppose that <math>\left|x+y\right|+\left|x-y\right|=2</math>. What is the maximum possible value of <math>x^2-6x+y^2</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
  
 
<math>
 
<math>
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\textbf{(E)}\ 9 </math>
 
\textbf{(E)}\ 9 </math>
  
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== 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>\boxed{\textbf{(D)}\ 8}</math>.
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== Solution 2 ==
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Since the equation <math>|x+y|+|x-y| = 2</math> is dealing with absolute values, the following 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>, we care most about <math>-6x,</math> since both <math>x^2</math> and <math>y^2</math> are non-negative. 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{\textbf{(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}}

Latest revision as of 20:50, 2 November 2021

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

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

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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