AMC 10 · 2004 · #15
Grade 7 algebraGiven that −4≤x≤−2 and 2≤y≤4, what is the largest possible value of xx+y?
Pick an answer.
AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: The number $x$ can be anywhere from $-4$ to $-2$, and the number $y$ can be anywhere from $2$ to $4$. Among all allowed pairs, find the largest value the expression $\frac{x+y}{x}$ can reach.
Givens: $x$ ranges over $-4 \leq x \leq -2$, so $x$ is always negative.; $y$ ranges over $2 \leq y \leq 4$, so $y$ is always positive.; The expression to maximize is $\frac{x+y}{x}$.; Answer choices: (A) $-1$, (B) $-\frac12$, (C) $0$, (D) $\frac12$, (E) $1$
Unknowns: The largest possible value of $\frac{x+y}{x}$ over the allowed ranges of $x$ and $y$.
Understand
Restated: The number $x$ can be anywhere from $-4$ to $-2$, and the number $y$ can be anywhere from $2$ to $4$. Among all allowed pairs, find the largest value the expression $\frac{x+y}{x}$ can reach.
Givens: $x$ ranges over $-4 \leq x \leq -2$, so $x$ is always negative.; $y$ ranges over $2 \leq y \leq 4$, so $y$ is always positive.; The expression to maximize is $\frac{x+y}{x}$.; Answer choices: (A) $-1$, (B) $-\frac12$, (C) $0$, (D) $\frac12$, (E) $1$
Plan
Primary tool: #14 Extreme Principle
Secondary: #15 Organize Information in More Ways, #3 Eliminate Possibilities
The question asks for a maximum over a range of inputs, which is exactly what the Extreme Principle (Tool #14) is for: the best value lives at a boundary corner, not somewhere in the middle. But before hunting corners, rewrite the messy fraction into a cleaner shape (Tool #15): $\frac{x+y}{x}=1+\frac{y}{x}$. That split makes the whole problem about one small piece, $\frac{y}{x}$, so it is obvious which way to push $x$ and $y$. Because $x<0$ and $y>0$, the piece $\frac{y}{x}$ is always negative, so maximizing means making it the least negative — the smallest $y$ over the largest $|x|$. Tool #3 (Eliminate Possibilities) then confirms the corner is a true maximum by comparing it against the opposite corner, which produces the smallest listed value.
Execute — Answer: D
6.EE.A.3 Step 1 Rewrite the fraction
- Split the fraction by dividing each part of the top by the bottom: $\frac{x+y}{x}=\frac{x}{x}+\frac{y}{x}=1+\frac{y}{x}$.
- The $1$ never changes, so the whole expression is as large as possible exactly when the leftover piece $\frac{y}{x}$ is as large as possible.
- This turns a two-letter fraction into a question about a single quantity, $\frac{y}{x}$.
💡 Separating out the fixed $1$ leaves just one moving part to worry about.
7.NS.A.2 Step 2 Read the signs
- In the allowed ranges, $x$ is always negative and $y$ is always positive.
- A positive number divided by a negative number is negative, so $\frac{y}{x}$ is always negative, no matter which allowed values you pick.
- That means $1+\frac{y}{x}$ is always less than $1$.
- To make it as big as possible, push $\frac{y}{x}$ up toward zero — that is, make the negative number $\frac{y}{x}$ as close to $0$ as it can get.
💡 Opposite signs on top and bottom force the quotient to be negative, so the best you can do is get it near zero.
7.NS.A.3 Step 3 Push to the best corner
- The size of $\frac{y}{x}$ is $\frac{y}{|x|}$.
- To make this fraction small (so the negative value is closest to zero), pick the smallest top and the largest bottom: the smallest $y$ is $2$ and the largest $|x|$ is $4$, which means $x=-4$.
- Then $\frac{y}{x}=\frac{2}{-4}=-\frac12$, so the expression is $1+\left(-\frac12\right)=\frac12$.
- That is the largest value, matching choice (D).
💡 A negative fraction shrinks toward zero when its top is small and its bottom is large.
6.EE.A.3 Split the fraction by dividing each part of the top by the bottom: $\frac{x+y}{x 7.NS.A.2 In the allowed ranges, $x$ is always negative and $y$ is always positive. A posi 7.NS.A.3 The size of $\frac{y}{x}$ is $\frac{y}{|x|}$. To make this fraction small (so th Review
Reasonableness: The result $\frac12$ is less than $1$, which fits: since $\frac{y}{x}$ is always negative, $1+\frac{y}{x}$ can never reach $1$ (choice (E)), so (E) is impossible and $\frac12$ being just under it looks right. The opposite corner is a good sanity check for the maximum: taking the largest $y=4$ over the smallest $|x|$, $x=-2$, gives $1+\frac{4}{-2}=1-2=-1$, which is choice (A) — the smallest listed value. So the expression really does swing from $-1$ up to $\frac12$, and $\frac12$ is the top.
Alternative: Skip the rewrite and just test the four corners of the rectangle directly (a make-a-list approach): $(x,y)=(-4,2)\to\frac{-2}{-4}=\frac12$, $(-4,4)\to\frac{0}{-4}=0$, $(-2,2)\to\frac{0}{-2}=0$, $(-2,4)\to\frac{2}{-2}=-1$. The largest of these is $\frac12$. Since the expression changes steadily as $x$ and $y$ move, the maximum must sit at a corner, so checking the four corners is enough and again gives (D).
CCSS standards used (min grade 7)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Rewriting $\frac{x+y}{x}$ as the equivalent expression $1+\frac{y}{x}$ to isolate the one varying piece.)7.NS.A.2Apply and extend understanding of multiplication and division of rational numbers (Recognizing that a positive $y$ over a negative $x$ makes $\frac{y}{x}$ negative, so the expression stays below $1$.)7.NS.A.3Solve real-world problems involving the four operations with rational numbers (Computing $\frac{2}{-4}=-\frac12$ at the best corner and combining it with $1$ to get $\frac12$.)
⭐ Split a hard fraction into a fixed part plus a moving part, then push the moving part to its extreme; with opposite signs on top and bottom, the quotient is smallest in size when the top is small and the bottom is big.
⭐ Split a hard fraction into a fixed part plus a moving part, then push the moving part to its extreme; with opposite signs on top and bottom, the quotient is smallest in size when the top is small and the bottom is big.
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