Difference between revisions of "1980 AHSME Problems/Problem 22"

(Solution)
Line 10: Line 10:
  
 
== Solution ==
 
== Solution ==
The first two given functions intersect at <math>\left(\frac{1}{3},\frac{7}{3}\right)</math>, and last two at <math>\left(\frac{2}{3},\frac{8}{3}\right)</math>. Therefore <cmath>f(x)=\left\{ \begin{matrix} 4x+1 & x<\frac{1}{3} \\
+
Let <math>a(x) = 4x+1</math>, <math>b(x) = x+2</math>, and <math>c(x) = -2x+4</math>.  Then <math>f(x) = \min(a(x), b(x), c(x))</math>.
                                          x+2   & \frac{1}{3}>x>\frac{2}{3} \\
+
 
                                          -2x+4 & x>\frac{2}{3}
+
<math>a(x) - b(x) = 3x - 1</math>, so <math>a(x) > b(x)</math> when <math>x > \frac{1}{3}</math>, <math>a(x) = b(x)</math> when <math>x = \frac{1}{3}</math>, and <math>a(x) < b(x)</math> when <math>x < \frac{1}{3}</math>.
\end{matrix}\right.
+
 
</cmath>
+
<math>a(x) - c(x) = 6x - 3</math>, so <math>a(x) > c(x)</math> when <math>x > \frac{1}{2}</math>, <math>a(x) = c(x)</math> when <math>x = \frac{1}{2}</math>, and <math>a(x) < c(x)</math> when <math>x < \frac{1}{2}</math>.
Which attains a maximum at
+
 
<math>\boxed{(E)\ \frac{8}{3}}</math>
+
<math>b(x) - c(x) = 3x - 2</math>, so <math>b(x) > c(x)</math> when <math>x > \frac{2}{3}</math>, <math>b(x) = c(x)</math> when <math>x = \frac{2}{3}</math>, and <math>b(x) < c(x)</math> when <math>x < \frac{2}{3}</math>.
 +
 +
If <math>x \leq \frac{1}{3}</math>, <math>a(x) \leq b(x) < c(x)</math>, so <math>f(x) = a(x) = 4x+1</math>.
 +
 
 +
If <math>\frac{1}{3} \leq x \leq \frac{1}{2}</math>, <math>b(x) \leq a(x) \leq c(x)</math>, so <math>f(x) = b(x) = x+2</math>.
 +
 
 +
If <math>\frac{1}{2} \leq x \leq \frac{2}{3}</math>, <math>b(x) \leq c(x) \leq a(x)</math>, so <math>f(x) = b(x) = x+2</math>.
 +
 
 +
If <math>x \geq \frac{2}{3}</math>, <math>c(x) \leq b(x) < a(x)</math>, so <math>f(x) = c(x) = -2x+4</math>.
 +
 
 +
So <math>f(x) </math> is increasing while <math>x < \frac{2}{3}</math> and decreasing while <math>x > \frac{2}{3}</math>, and therefore, <math>f(x)</math> is maximized at <math>x = \frac{2}{3}</math>, with a value of <math>f\left(\frac{2}{3}\right) = \frac{2}{3} + 2 = \boxed{(\textbf{E}) \frac{8}{3}}</math>
  
 
== See also ==
 
== See also ==

Revision as of 14:28, 16 August 2025

Problem

For each real number $x$, let $f(x)$ be the minimum of the numbers $4x+1, x+2$, and $-2x+4$. Then the maximum value of $f(x)$ is

$\text{(A)} \ \frac{1}{3} \qquad  \text{(B)} \ \frac{1}{2} \qquad  \text{(C)} \ \frac{2}{3} \qquad  \text{(D)} \ \frac{5}{2} \qquad  \text{(E)}\ \frac{8}{3}$

Solution

Let $a(x) = 4x+1$, $b(x) = x+2$, and $c(x) = -2x+4$. Then $f(x) = \min(a(x), b(x), c(x))$.

$a(x) - b(x) = 3x - 1$, so $a(x) > b(x)$ when $x > \frac{1}{3}$, $a(x) = b(x)$ when $x = \frac{1}{3}$, and $a(x) < b(x)$ when $x < \frac{1}{3}$.

$a(x) - c(x) = 6x - 3$, so $a(x) > c(x)$ when $x > \frac{1}{2}$, $a(x) = c(x)$ when $x = \frac{1}{2}$, and $a(x) < c(x)$ when $x < \frac{1}{2}$.

$b(x) - c(x) = 3x - 2$, so $b(x) > c(x)$ when $x > \frac{2}{3}$, $b(x) = c(x)$ when $x = \frac{2}{3}$, and $b(x) < c(x)$ when $x < \frac{2}{3}$.

If $x \leq \frac{1}{3}$, $a(x) \leq b(x) < c(x)$, so $f(x) = a(x) = 4x+1$.

If $\frac{1}{3} \leq x \leq \frac{1}{2}$, $b(x) \leq a(x) \leq c(x)$, so $f(x) = b(x) = x+2$.

If $\frac{1}{2} \leq x \leq \frac{2}{3}$, $b(x) \leq c(x) \leq a(x)$, so $f(x) = b(x) = x+2$.

If $x \geq \frac{2}{3}$, $c(x) \leq b(x) < a(x)$, so $f(x) = c(x) = -2x+4$.

So $f(x)$ is increasing while $x < \frac{2}{3}$ and decreasing while $x > \frac{2}{3}$, and therefore, $f(x)$ is maximized at $x = \frac{2}{3}$, with a value of $f\left(\frac{2}{3}\right) = \frac{2}{3} + 2 = \boxed{(\textbf{E}) \frac{8}{3}}$

See also

1980 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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 26 27 28 29 30
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png