Difference between revisions of "2004 AMC 10A Problems/Problem 15"
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==Problem== | ==Problem== | ||
Given that <math>-4\leq x\leq-2</math> and <math>2\leq y\leq4</math>, what is the largest possible value of <math>\frac{x+y}{x}</math>? | Given that <math>-4\leq x\leq-2</math> and <math>2\leq y\leq4</math>, what is the largest possible value of <math>\frac{x+y}{x}</math>? | ||
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-JinhoK | -JinhoK | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/LBgCCFdCvYc | ||
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+ | Education, the Study of Everything | ||
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==See also== | ==See also== |
Revision as of 13:18, 21 April 2021
Problem
Given that and , what is the largest possible value of ?
Solution
Rewrite as .
We also know that because and are of opposite sign.
Therefore, is maximized when is minimized, which occurs when is the largest and is the smallest.
This occurs at , so .
Solution 2
If the answer choice is valid, then it must satisfy . We use answer choices from greatest to least since the question asks for the greatest value.
Answer choice . We see that if then
and . However, is not in the domain of , so is incorrect.
Answer choice , however, we can find a value that satisfies which simplifies to , such as .
Therefore, is the greatest.
Solution 3
As , we know that the denominator of our given fraction is negative. So to achieve the greatest value possible, our numerator, or must also be as small as possible.
So we pick our smallest value for , which is .
Now if we if set our value of to its lowest, our expression becomes As we decrease our value, we see that our numerator decrease from and our denominator decrease from at the same rate. So decreasing our value decreases the overwhelming gap between our denominator and numerator, which gives us an overall bigger number.
So we also pick the smallest value for , which is . We know have .
Therefore, is our greatest possible value.
-JinhoK
Video Solution
Education, the Study of Everything
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.