If $A, B, C,$ and $D$ are positive numbers such that $A + 2B + 3C + 4D = 8$, then what is the maximum value of $ABCD$?
Any help would be greatly appreciated. Thanks!
[Math] If $A, B, C,$ and $D$ are positive numbers such that $A + 2B + 3C + 4D = 8$, then what is the maximum value of $ABCD$
inequalitymaxima-minima
Best Answer
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 $\frac{2}{3}$