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Difference between revisions of "1982 AHSME Problems/Problem 29"

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Recall that <math>x_0 \leq 1-3x_0 \leq 2x_0,</math> so <math>\frac15 \leq x_0 \leq \frac14.</math> This problem is equivalent to finding the minimum value of <cmath>f(x)=xyz=x(1-3x)(2x)=2x^2(1-3x)</cmath> in the interval <math>I=\left[\frac15,\frac14\right].</math> The graph of <math>y=f(x)</math> is shown below:
 
Recall that <math>x_0 \leq 1-3x_0 \leq 2x_0,</math> so <math>\frac15 \leq x_0 \leq \frac14.</math> This problem is equivalent to finding the minimum value of <cmath>f(x)=xyz=x(1-3x)(2x)=2x^2(1-3x)</cmath> in the interval <math>I=\left[\frac15,\frac14\right].</math> The graph of <math>y=f(x)</math> is shown below:
−
............DIAGRAM GOES HERE ............
+
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(900,200);
 +
 
 +
real f(real x) { return 2x^2 * (1-3x); }
 +
 
 +
real xMin = -0.35;
 +
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 <math>f</math> has a relative minimum at <math>x=0,</math> and cubic functions have at most one relative minimum, we conclude that the minimum value of <math>f</math> in <math>I</math> is at either <math>x=\frac15</math> or <math>x=\frac14.</math> As <math>f\left(\frac14\right)=\frac{1}{32}\leq f\left(\frac15\right)=\frac{4}{125},</math> the minimum value of <math>f</math> in <math>I</math> is <math>\boxed{\textbf{(A)}\ \frac{1}{32}}.</math>
 
Since <math>f</math> has a relative minimum at <math>x=0,</math> and cubic functions have at most one relative minimum, we conclude that the minimum value of <math>f</math> in <math>I</math> is at either <math>x=\frac15</math> or <math>x=\frac14.</math> As <math>f\left(\frac14\right)=\frac{1}{32}\leq f\left(\frac15\right)=\frac{4}{125},</math> the minimum value of <math>f</math> in <math>I</math> is <math>\boxed{\textbf{(A)}\ \frac{1}{32}}.</math>
  

Revision as of 22:16, 17 September 2021

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 = -0.35; 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 (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
All AHSME Problems and Solutions


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