2004 AMC 10A Problem 15

P11-15~3 minPrint

Absolute Value & Inequalities Algebraic Manipulation Bound the quantity above/below or estimate to pin it down Consider the largest/smallest element or boundary case

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

Show solution without answering

Key insight

Split the fraction into 1 + y/x. Since y/x is always negative, make it as close to 0 as possible: y = 2 and x = -4 give 1/2.

Solution

1. The fraction is easier to think about if you split it:
$\frac{x+y}{x} = \frac{x}{x} + \frac{y}{x} = 1 + \frac{y}{x}.$
2. Here $y$ is always positive and $x$ is always negative, so $\dfrac{y}{x}$ is always negative. The whole expression is $1$ minus something.
3. To make $1$ minus something as big as possible, make that something as small as possible. The size of $\dfrac{y}{x}$ is $y$ divided by the size of $x$ . That is smallest when $y$ is as small as possible and $x$ is as far from $0$ as possible.
4. So take $y = 2$ and $x = -4$ :
$1 + \frac{2}{-4} = 1 - \frac12 = \frac12.$

The answer is $\boxed{\textbf{(D)}\ \frac12}$ .

Why this works

Rewriting the fraction as $1 + \frac{y}{x}$ separates the fixed part from the part that changes, and makes it clear which way each variable should go. For a fraction like this, the largest and smallest values happen when each variable is at one end of its allowed range, so the corners are the places to look.

Alternative approach

Try all four corners, where each variable is at one end of its range:

$x$	$y$	$\frac{x+y}{x}$
$-4$	$2$	$\frac{-2}{-4} = \frac12$
$-4$	$4$	$\frac{0}{-4} = 0$
$-2$	$2$	$\frac{0}{-2} = 0$
$-2$	$4$	$\frac{2}{-2} = -1$

The largest is $\tfrac12$ . Choice (E), $1$ , cannot happen, because it would need $y = 0$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Split the fraction:
$\frac{x+y}{x} = 1 + \frac{y}{x}.$
Since $y > 0$ and $x < 0$ , the term $\frac{y}{x}$ is always negative, so the expression is always less than $1$ . To make it as large as possible we want $\frac{y}{x}$ as close to $0$ as possible, i.e. $\left|\frac{y}{x}\right| = \frac{y}{|x|}$ as small as possible.

That means the smallest $y$ and the largest $|x|$ : $y = 2$ and $x = -4$ . Then
$1 + \frac{2}{-4} = 1 - \frac12 = \frac12.$

The answer is $\boxed{\textbf{(D)}\ \frac12}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 20: Special Functions (§20.2, §20.5)
Introduction to Algebra, Chapter 11: Special Factorizations (§11.2, §11.3, §11.5)
Introduction to Algebra, Chapter 15: More Inequalities (§15.1, §15.3, §15.4)
Introduction to Algebra, Chapter 22: Special Manipulations