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

(Created page with "== Problem == For each real number <math>x</math>, let <math>f(x)</math> be the minimum of the numbers <math>4x+1, x+2</math>, and <math>-2x+4</math>. Then the maximum value of ...")
 
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== Solution ==
 
== Solution ==
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<math>\fbox{}</math>
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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>.
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<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>.
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<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>.
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<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>.
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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>.
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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>.
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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>.
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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>.
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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 ==

Latest 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 (Problems • Answer Key • Resources)
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

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