1982 AMC 12 Problem 29

Attempt Problem 29 of the 1982 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1982 AMC 12 solutions, or check the answer key.

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29.

Let x,x, y,y, and zz be three positive real numbers whose sum is 1.1. If no one of these numbers is more than twice any other, then the minimum possible value of the product xyzxyz is
132\frac1{32}
136\frac1{36}
4125\frac4{125}
1127\frac1{127}
none of these
Answer: A
Concepts:optimizationinequalityextremal argument
Small hint:
Order the variables x≤y≤zx\le y\le z; then the active constraint is z≤2xz\le2x
Big hint:
A minimum occurs on the boundary z=2x,z=2x, where y=1−3xy=1-3x
Solution:
Order x≤y≤z.x\le y\le z. At a minimum the spread is maximal, so z=2xz=2x and y=1−3x.y=1-3x. The ordering requires 15≤x≤14.\frac{1}{5}\le x\le\frac{1}{4}. Thus xyz=2x2(1−3x).xyz=2x^2(1-3x). Its only interior critical point is a maximum, so compare endpoints: the values are 4125\frac{4}{125} and 132,\frac{1}{32}, respectively. The minimum is 132.\frac{1}{32}.
Therefore, the correct answer is A.
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