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

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==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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<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>
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==Solution==
 
The answer is A, 1/32, as obtained by (1/4) * (1/4) * (1/2).
 
The answer is A, 1/32, as obtained by (1/4) * (1/4) * (1/2).
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==See Also==
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{{AHSME box|year=1982|num-b=28|num-a=30}}
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{{MAA Notice}}

Latest revision as of 16:41, 17 June 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

The answer is A, 1/32, as obtained by (1/4) * (1/4) * (1/2).

See Also

1982 AHSME (ProblemsAnswer KeyResources)
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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