1951 AMC 12 Problem 43

Attempt Problem 43 of the 1951 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 1951 AMC 12 solutions, or check the answer key.

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43.

Of the following statements, the only one that is incorrect is:

An inequality will remain true after each side is increased, decreased, multiplied, or divided (zero excluded) by the same positive quantity.

The arithmetic mean of two unequal positive quantities is greater than their geometric mean.

If the sum of two positive quantities is given, their product is largest when they are equal.

If aa and bb are positive and unequal, 12(a2+b2)\dfrac12(a^2+b^2) is greater than [12(a+b)]2.\left[\dfrac12(a+b)\right]^2.

If the product of two positive quantities is given, their sum is greatest when they are equal.

Answer: E
Concepts:AM-GM Inequalityoptimization
Difficulty rating: 1450
Small Hint:

For a fixed positive product, compare the sum at equality with sums from increasingly unequal factors

Big Hint:

AM-GM gives a minimum, not a maximum, for the sum when the product is fixed

Solution:

If uv=P>0,uv=P\gt0, then u+v2P,u+v\ge2\sqrt P, with equality at u=v.u=v. Thus equality gives the least possible sum. The sum can grow without bound by taking uu large and v=Puv=\frac{P}{u} small, so it is not greatest at equality.

Thus, the correct answer is E.

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Problem 43 in Other Years

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