Difference between revisions of "2004 AMC 10A Problems/Problem 15"
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Answer choice <math>\text{(D)}</math>, however, we can find a value that satisfies <math>\frac{x+y}{x}=\frac{1}{2}</math> which simplifies to <math>x+2y=0</math>, such as <math>(-4,2)</math>. | Answer choice <math>\text{(D)}</math>, however, we can find a value that satisfies <math>\frac{x+y}{x}=\frac{1}{2}</math> which simplifies to <math>x+2y=0</math>, such as <math>(-4,2)</math>. | ||
| − | Therefore, <math>\ | + | Therefore, <math>\framebox{\text{(D)}</math> is the greatest. |
==See also== | ==See also== | ||
Revision as of 11:45, 14 December 2019
Contents
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, $\framebox{\text{(D)}$ (Error compiling LaTeX. Unknown error_msg) is the greatest.
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. 
