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Difference between revisions of "1982 AHSME Problems/Problem 29"
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==Problem==
 
==Problem==
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Let <math>x,y</math>, and <math>z</math> be three positive real numbers whose sum is <math>1</math>. If no one of these numbers is more than twice any other, then the minimum possible value of the product <math>xyz</math> is
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Let <math>x,y</math>, and <math>z</math> be three positive real numbers whose sum is <math>1</math>. If no one of these numbers is more than twice any other,  
 +
then the minimum possible value of the product <math>xyz</math> is  
 +
 
 +
<math> \textbf{(A)}\ \frac{1}{32}\qquad
 +
\textbf{(B)}\ \frac{1}{36}\qquad
 +
\textbf{(C)}\ \frac{4}{125}\qquad
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\textbf{(D)}\ \frac{1}{127}\qquad
 +
\textbf{(E)}\ \text{none of these} </math> 
  
−
<math>\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}</math>
 
 
==Solution==
 
==Solution==
−
The answer is A, 1/32, as obtained by (1/4) * (1/4) * (1/2).
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Suppose that the product <math>xyz</math> is minimized at <math>(x,y,z)=(x_0,y_0,z_0).</math> Without the loss of generality, let <math>x_0 \leq y_0 \leq z_0</math> and fix <math>y=y_0.</math>
 +
 
 +
To minimize <math>xy_0z,</math> we minimize <math>xz.</math> Note that <math>x+z=1-y_0.</math> By a corollary of the AM-GM Inequality <i><b>(If two nonnegative numbers have a constant sum, then their product is minimized when they are as far as possible.)</b></i>, we get <math>z_0=2x_0.</math> It follows that <math>y_0=1-3x_0.</math>
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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> Since <math>f(x)</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(x)</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(x)</math> in <math>I</math> is <math>\boxed{\textbf{(A)}\ \frac{1}{32}}.</math>
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~MRENTHUSIASM
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==See Also==
 
==See Also==
 
{{AHSME box|year=1982|num-b=28|num-a=30}}
 
{{AHSME box|year=1982|num-b=28|num-a=30}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:13, 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 $I=\left[\frac15,\frac14\right].$ Since $f(x)$ has a relative minimum at $x=0,$ and cubic functions have at most one relative minimum, we conclude that the minimum value of $f(x)$ 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 $f(x)$ 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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