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If w, x, y, and z are positive real numbers such that w + 2x + 3y + 4z = 8, then what is the maximum value of wxyz?

 Aug 14, 2021
 #1
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The minimum value is 9/25.

 Aug 14, 2021
 #2
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By AM-GM,

\(\frac{w+2x+3y+4z}{4} \ge \sqrt[4]{24wxyz}\\ 2\ge\sqrt[4]{24wxyz}\\ 16\ge24wxyz\\ \frac{2}{3}\ge wxyz\)

so the maximum value is \(\boxed{\frac{2}{3}}\)

for completeness, equality occurs when the terms are equal: \(w=2, x=1, y=\frac{2}{3}, z=\frac{1}{2}\)

 Aug 21, 2021

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