2004 AMC 10A Problem 15

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 $\boxed{\textbf{(D)}\ \frac12}$.

Dividing through by $x$ isolates the only part that varies, $\frac{y}{x}$, whose sign is fixed on the given box. A quotient of positive quantities is monotone in each variable separately, so its extreme values occur at corners of the domain. Look for a rewrite that makes the monotonicity obvious before testing points.

Test the four corners of the rectangle: $(x, y) = (-4, 2)$ gives $\tfrac12$; $(-4, 4)$ gives $0$; $(-2, 2)$ gives $0$; $(-2, 4)$ gives $-1$. The largest is $\tfrac12$. Choice (E) is impossible since $\frac{x+y}{x} = 1$ would force $y = 0$.

• Maximizing the numerator x + y and the denominator separately, or plugging in x = -2, y = 2 to get 0.
• Forgetting that dividing by a negative $x$ flips the direction of inequalities, and picking the "largest" $x = -2$.