Difference between revisions of "1982 AHSME Problems/Problem 29"
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| + | ==Problem== | ||
| + | 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> | ||
| + | ==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). | ||
| + | ==See Also== | ||
| + | {{AHSME box|year=1982|num-b=28|num-a=30}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Revision as of 15:41, 17 June 2021
Problem
Let 
, and 
be three positive real numbers whose sum is 
. If no one of these numbers is more than twice any other, then the minimum possible value of the product 
is
Solution
The answer is A, 1/32, as obtained by (1/4) * (1/4) * (1/2).
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
