1982 AHSME Problems/Problem 29

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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: [asy] /* Made by MRENTHUSIASM */ size(900,200);   real f(real x) { return 2x^2 * (1-3x); }   real xMin = -0.349; real xMax = 1/2; real yMin = -1/4; real yMax = 1/2;  draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2));  pair A[]; A[0] = (0,0); A[1] = (1/2,1/2); A[2] = (2/3,1/3);  draw((1/5,-0.015)--(1/5,0.015),linewidth(1)); draw((1/4,-0.015)--(1/4,0.015),linewidth(1)); draw((1/3,-0.015)--(1/3,0.015),linewidth(1));  draw((1/5,yMin)--(1/5,yMax),dashed); draw((1/4,yMin)--(1/4,yMax),dashed);  label("$0$",A[0],(-2,-2)); label("$\frac15$",(1/5,0),(0,-2),UnFill); label("$\frac14$",(1/4,0),(0,-2),UnFill); label("$\frac13$",(1/3,0),(0,-2));  draw(graph(f,xMin,xMax),red); dot(A[0],red+linewidth(5));  dot((1/3,0),red+linewidth(5)); [/asy] 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

See Also

1982 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 28
Followed by
Problem 30
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All AHSME Problems and Solutions


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