1982 AHSME Problems/Problem 29

Problem

Let $x,y$ , and $z$ be three positive real numbers whose sum is $1$ . If no one of these numbers is more than twice any other, then the minimum possible value of the product $xyz$ is

$\textbf{(A)}\ \frac{1}{32}\qquad \textbf{(B)}\ \frac{1}{36}\qquad \textbf{(C)}\ \frac{4}{125}\qquad \textbf{(D)}\ \frac{1}{127}\qquad \textbf{(E)}\ \text{none of these}$

Solution

Suppose that the product $xyz$ is minimized at $(x,y,z)=(x_0,y_0,z_0).$ Without the loss of generality, let $x_0 \leq y_0 \leq z_0$ and fix $y=y_0.$

To minimize $xy_0z,$ we minimize $xz.$ Note that $x+z=1-y_0.$ By a corollary of the AM-GM Inequality (If two nonnegative numbers have a constant sum, then their product is minimized when they are as far as possible.), we get $z_0=2x_0.$ It follows that $y_0=1-3x_0.$

Recall that $x_0 \leq 1-3x_0 \leq 2x_0,$ so $\frac15 \leq x_0 \leq \frac14.$ This problem is equivalent to finding the minimum value of $f(x)=xyz=x(1-3x)(2x)=2x^2(1-3x)$ in the interval $I=\left[\frac15,\frac14\right].$ The graph of $y=f(x)$ is shown below: ............DIAGRAM GOES HERE ............ Since $f$ has a relative minimum at $x=0,$ and cubic functions have at most one relative minimum, we conclude that the minimum value of $f$ in $I$ is at either $x=\frac15$ or $x=\frac14.$ As $f\left(\frac14\right)=\frac{1}{32}\leq f\left(\frac15\right)=\frac{4}{125},$ the minimum value of $f$ in $I$ is $\boxed{\textbf{(A)}\ \frac{1}{32}}.$

~MRENTHUSIASM

1982 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 28	Followed by Problem 30
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
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1982 AHSME Problems/Problem 29

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