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
Difficulty rating: 2340
Small Hint:

Order the variables xyzx\le y\le z; then the active constraint is z2xz\le2x

Big Hint:

A minimum occurs on the boundary z=2x,z=2x, where y=13xy=1-3x

Solution:

Order xyz.x\le y\le z. At a minimum the spread is maximal, so z=2xz=2x and y=13x.y=1-3x. The ordering requires 15x14.\frac{1}{5}\le x\le\frac{1}{4}. Thus xyz=2x2(13x).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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