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# Inequality

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Let w, x, y, and z be positive real numbers.  If w + x + y + z = 8, then what is the maximum value of wxyz?

Oct 15, 2021

### 1+0 Answers

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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

Oct 16, 2021