Let w, x, y, and z be positive real numbers. If w + x + y + z = 8, then what is the maximum value of wxyz?
+678 The maximum value for \(wxyz\), where it applies \(w+x+y+z=8\) ,
is 16 (\(2+2+2+2=8\)) and (\(2*2*2*2=16\)).
Let's look, how I did this:
If \(w\) (or \(x,y,z\)) is 8 and the others are only 0, this wouldn't be possible because 8*0*0*0 = 8,
If \(w\) (or \(x,y,z\)) is 7 can't stand 0,7 or like that because that makes the number smaller and isn't a real number.
So 7*1*0*0 is still 0 because if at least one factor a zero, then the product is ALWAYS 0,
so we have to distribute the factors with integers that are also not 0. After this rule, we're starting now with 4:
(by 5 and 6 is the product 0, so it's not working there)
4*1*1*2 = 8 ( - too small),
let us look by 3:
3*2*1*2 = 12 ( - too small for 16) or other possibilities would be the product (\(n<13\)).
By 2 it works great:
(\(2^4\)) or (\(2*2*2*2 \)) eqauls 16. There is no possibility that equals 16, except one.
By 1 isn't it possible (reason: The product of \(1*x*y*z\) is maximum \(n<13\) ).
So if you don't know what's the maximum of \(w*x*y*z\) is 16 (\(2*2*2*2\) ).
Straight
