Difference between revisions of "2004 AMC 10A Problems/Problem 15"
(→Solution) |
m (+box) |
||
Line 1: | Line 1: | ||
==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 | + | 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>? |
<math> \mathrm{(A) \ } -1 \qquad \mathrm{(B) \ } -\frac12 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac12 \qquad \mathrm{(E) \ } 1 </math> | <math> \mathrm{(A) \ } -1 \qquad \mathrm{(B) \ } -\frac12 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac12 \qquad \mathrm{(E) \ } 1 </math> | ||
Line 13: | Line 13: | ||
This occurs at (-4,2), so <math>\frac{x+y}x=1-\frac12=\frac12\Rightarrow \mathrm{(D)}</math>. | This occurs at (-4,2), so <math>\frac{x+y}x=1-\frac12=\frac12\Rightarrow \mathrm{(D)}</math>. | ||
− | ==See | + | ==See also== |
− | + | {{AMC10 box|year=2004|ab=A|num-b=14|num-a=16}} | |
− | |||
− | |||
− | |||
− | |||
− | |||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Revision as of 21:00, 28 February 2007
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 (-4,2), so .
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 |