Difference between revisions of "1983 AIME Problems/Problem 2"
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== Problem == | == Problem == | ||
Let <math>f(x)=|x-p|+|x-15|+|x-p-15|</math>, where <math>0 < p < 15</math>. Determine the [[minimum]] value taken by <math>f(x)</math> for <math>x</math> in the [[interval]] <math>p \leq x\leq15</math>. | Let <math>f(x)=|x-p|+|x-15|+|x-p-15|</math>, where <math>0 < p < 15</math>. Determine the [[minimum]] value taken by <math>f(x)</math> for <math>x</math> in the [[interval]] <math>p \leq x\leq15</math>. | ||
== Solution == | == Solution == | ||
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| + | === Solution 1 === | ||
It is best to get rid of the [[absolute value]]s first. | It is best to get rid of the [[absolute value]]s first. | ||
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Adding these together, we find that the sum is equal to <math>30-x</math>, which attains its minimum value (on the given interval <math>p \leq x \leq 15</math>) when <math>x=15</math>, giving a minimum of <math>\boxed{015}</math>. | Adding these together, we find that the sum is equal to <math>30-x</math>, which attains its minimum value (on the given interval <math>p \leq x \leq 15</math>) when <math>x=15</math>, giving a minimum of <math>\boxed{015}</math>. | ||
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| + | === Solution 2 === | ||
| + | Let <math>p</math> be equal to <math>15 - \varepsilon</math>, where <math>\varepsilon</math> is an almost neglectable value. Because of the small value <math>\varepsilon</math>, the [[domain]] of <math>f(x)</math> is basically the [[set]] <math>{15}</math>. plugging in <math>15</math> gives <math>\varepsilon + 0 + 15 - \varepsilon</math>, or <math>15</math>, so the answer is <math>\boxed{15}</math> | ||
== See Also == | == See Also == | ||
Latest revision as of 21:19, 20 October 2024
Problem
Let 
, where 
. Determine the minimum value taken by 
for 
in the interval 
.
Solution
Solution 1
It is best to get rid of the absolute values first.
Under the given circumstances, we notice that 
, 
, and 
.
Adding these together, we find that the sum is equal to 
, which attains its minimum value (on the given interval 
) when 
, giving a minimum of 
.
Solution 2
Let 
be equal to 
, where 
is an almost neglectable value. Because of the small value , the domain of is basically the set 
. plugging in 
gives 
, or , so the answer is 
See Also
| 1983 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
