Difference between revisions of "1983 AIME Problems/Problem 2"

(Problem)
(Solution)
 
(9 intermediate revisions by 7 users not shown)
Line 1: Line 1:
 +
__TOC__
 +
 
== Problem ==
 
== Problem ==
Let <math>f(x)=|x-p|+|x-15|+|x-p-15|</math>, where <math>p \leq x \leq 15</math>. Determine the [[minimum]] value taken by <math>f(x)</math> for <math>x</math> in the [[interval]] <math>p < 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 ==
It is best to get rid of the [[absolute value]] first.  
+
 
 +
=== Solution 1 ===
 +
It is best to get rid of the [[absolute value]]s first.  
  
 
Under the given circumstances, we notice that <math>|x-p|=x-p</math>, <math>|x-15|=15-x</math>, and <math>|x-p-15|=15+p-x</math>.
 
Under the given circumstances, we notice that <math>|x-p|=x-p</math>, <math>|x-15|=15-x</math>, and <math>|x-p-15|=15+p-x</math>.
  
Adding these together, we find that the sum is equal to <math>30-x</math>, of which the minimum value is attained when <math>x=\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>.
 
 
Edit: <math>|x-p-15|</math> can equal <math>15+p-x</math> or <math>x-p-15</math> (for example, if <math>x=7</math> and <math>p=-12</math>, <math>x-p-15=4</math>). Thus, our two "cases" are
 
<math>30-x</math> (if <math>x-p\leq15</math>) and <math>x-2p</math> (if <math>x-p\geq15</math>). However, both of these cases give us <math>15</math> as the minimum value for <math>f(x)</math>, which indeed is the answer posted above.
 
 
 
  
Also note the lowest value occurs when <math>x=p=15</math> because this make the first two requirements <math>0</math>. It is easy then to check that 15 is the minimum value.
+
=== 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 $f(x)=|x-p|+|x-15|+|x-p-15|$, where $0 < p < 15$. Determine the minimum value taken by $f(x)$ for $x$ in the interval $p \leq x\leq15$.

Solution

Solution 1

It is best to get rid of the absolute values first.

Under the given circumstances, we notice that $|x-p|=x-p$, $|x-15|=15-x$, and $|x-p-15|=15+p-x$.

Adding these together, we find that the sum is equal to $30-x$, which attains its minimum value (on the given interval $p \leq x \leq 15$) when $x=15$, giving a minimum of $\boxed{015}$.

Solution 2

Let $p$ be equal to $15 - \varepsilon$, where $\varepsilon$ is an almost neglectable value. Because of the small value $\varepsilon$, the domain of $f(x)$ is basically the set ${15}$. plugging in $15$ gives $\varepsilon + 0 + 15 - \varepsilon$, or $15$, so the answer is $\boxed{15}$

See Also

1983 AIME (ProblemsAnswer KeyResources)
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