If $A, B, C,$ and $D$ are positive numbers such that $A + 2B + 3C + 4D = 8$, then what is the maximum value of $ABCD$?
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2026-04-07 20:04:47.1775592287
If $A, B, C,$ and $D$ are positive numbers such that $A + 2B + 3C + 4D = 8$, then what is the maximum value of $ABCD$?
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By A.M.-G.M.,
$$ 8=A+2B+3C+4D\ge4\sqrt[4]{A(2B)(3C)(4D)}$$
$$24ABCD\le2^4$$
$$ABCD\le\frac{2}{3}$$
The equality holds when $A=2B=3C=4D=2$.
The maximum value is $\frac{2}{3}$