Question

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1.)If u + v + w + x + y = 15, then what is the maximum value of uvx + uvy + uwx + uwy?
Notice that

uvx+uvy+uwx+uwy=u(x+y)(v+w)

If u,v,w,x,y are allowed to be negative, there is no largest value, since, for example, one could let u,x,y be arbitrarily “large” negative numbers (so u(x+y) is positive) then v+w is a large positive number and the whole product can be as large as you like.
If they are restricted to positive real numbers, then from AM-GM, we have
$\frac{u+(x+y)+(v+w)}{3}$ ≥ $\sqrt[3]{u(x+y)(v+w)}$
Using the initial condition, the left side is just 5 , so we have found
u(x+y)(v+w) ≤ $5^{3}$ =125
The maximum value is 125 , and occurs if, and only if, u=x+y=v+w=5 .
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Answer 1

Answer:

Only assuming u,v,x,y all being positive and diferent integers, we want maximum value in uvx + uvy + uwx + uwy where u+v+w+x+y = 15 => u² = 15u-(uv+uw+ux+uy) so we can assume that when uvx + uvy + uwx + uwy is maximum (uv+uw+ux+uy) is maximum we call this value a so u²-15u+a= 0 => (u-8)(u-7)=0 so for this combination a= 56 when u= 7 or 8. So say, u=7 then 7(v+w+x+y) = 56 in decreasing integer order we can only put 4,3,1,0 and for u=8 we can't get the four v,w,x,y different. With this combination we say uvx + uvy + uwx + uwy = 7[4(1+0)+3(1+0)]= 49 which turns out right as the value zero is not multiplying.

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Answer 2

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