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Difference between revisions of "2004 AMC 10A Problems/Problem 15"

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== Solution 3==
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As <math>-4\leq x\leq-2</math>, we know that the denominator of our given fraction is negative. So to achieve the greatest value possible, our numerator, or <math>x+y</math> must also be as small as possible.
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So we pick our smallest value for <math>y</math>, which is <math>2</math>.
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Now if we if set our value of <math>x</math> to its lowest, our expression becomes <math>\frac{(-2+2)}{-2} = \frac{0}{2}</math>
  
 
== Solution 3==
 
== Solution 3==

Revision as of 07:11, 22 December 2020

Problem

Given that $-4\leq x\leq-2$ and $2\leq y\leq4$, what is the largest possible value of $\frac{x+y}{x}$?

$\mathrm{(A) \ } -1 \qquad \mathrm{(B) \ } -\frac12 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac12 \qquad \mathrm{(E) \ } 1$

Solution

Rewrite $\frac{(x+y)}x$ as $\frac{x}x+\frac{y}x=1+\frac{y}x$.

We also know that $\frac{y}x<0$ because $x$ and $y$ are of opposite sign.

Therefore, $1+\frac{y}x$ is maximized when $|\frac{y}x|$ is minimized, which occurs when $|x|$ is the largest and $|y|$ is the smallest.

This occurs at $(-4,2)$, so $\frac{x+y}x=1-\frac12=\frac12\Rightarrow\boxed{\mathrm{(D)}\ \frac{1}{2}}$.


Solution 2

If the answer choice is valid, then it must satisfy $\frac{(x+y)}x$. We use answer choices from greatest to least since the question asks for the greatest value.

Answer choice $\text{(E)}$. We see that if $\frac{(x+y)}x = 1$ then

$x+y=x$ and $y=0$. However, $0$ is not in the domain of $y$, so $\text{(E)}$ is incorrect.

Answer choice $\text{(D)}$, however, we can find a value that satisfies $\frac{x+y}{x}=\frac{1}{2}$ which simplifies to $x+2y=0$, such as $(-4,2)$.

Therefore, $\boxed{\text{(D)}}$ is the greatest.


Solution 3

As $-4\leq x\leq-2$, we know that the denominator of our given fraction is negative. So to achieve the greatest value possible, our numerator, or $x+y$ must also be as small as possible.

So we pick our smallest value for $y$, which is $2$.

Now if we if set our value of $x$ to its lowest, our expression becomes $\frac{(-2+2)}{-2} = \frac{0}{2}$

Solution 3

See also

2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 10 Problems and Solutions

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