1980 AMC 12 Problem 22

Attempt Problem 22 of the 1980 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1980 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

22.

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)

\frac13

\frac12

\frac23

\frac52

\frac83

Sketch the three lines and follow their lower envelope

The maximum occurs where the active increasing line meets the decreasing line

The lower envelope is

4x+1

until it meets

x+2

x=\frac{1}{3},

then is

x+2

until that line meets

-2x+4

x=\frac{2}{3},

and thereafter is

-2x+4.

Thus its maximum occurs at

x=\frac{2}{3}

and equals

\frac23+2=\frac83.

Therefore, the correct answer is E.

Problem 22 in other years