Maximize Product of 3 Non-Negative Summands

[Telemachus]

Well, I have this multivariable calculus optimization problem. It says: Decompose a positive number on three non negative summands so that the product of them is maximum.

I thought of something like
$$w=x+y+z$$, $$w>0, x \geq{}0 , y \geq{}0 , z \geq{}0$$
$$f(x,y,z)=xyz$$
$$f_x=yz,f_y=xz,f_z=xy$$
The thing is I don't see any maximum here, clearly I'm not setting the things right.

 

[ehild]

F(x,y,z)=xyz has to be maximum with the condition G(x,y,z)= x+y+z - w=0. This conditional maximum can be found by the method of Lagrange multiplier, multiplying the condition with a parameter L and finding the possible extrema of the new function F-LG=xyz-L(x+y+z - w). You get three equation from setting the partial derivatives equal to zero, the forth equation is the condition G=0.



ehild

 

[Telemachus]

Thanks!

 

[Telemachus]

Mmm I think I didn't get it. Let's see what's wrong. I haven't used lagrange multipliers yet, but I want to follow your indications anyway, cause I'll have to use it later. So, It would be great if you can guide me with this, and tell me what I'm doing wrong.

$$w=x+y+z\Rightarrow{G=x+y+z-w=0}$$
$$F=xyz$$
$$F-\lambda G=xyz-\lambda(x+y+z-w)$$

$$\begin{Bmatrix}{ f_x=yz-\lambda=0\\f_y=xz-\lambda=0\\f_z=xy-\lambda=0\\x+y+z-w=0 \end{matrix}$$
$$\begin{Bmatrix}{ y=\displaystyle\frac{\lambda}{z}\\x=\displaystyle\frac{\lambda}{z}\end{matrix}$$
$$\Longrightarrow{xy=\lambda\rightarrow{\displaystyle\frac{\lambda^2}{z^2}=\lambda\rightarrow{z=\sqrt[ ]{\lambda}}}}$$
$$w=3\sqrt[ ]{\lambda}$$
Which I think its inconclusive, I'm think I'm not using the Lagrange multipliers on the right way, I don't know how to, so if I'm not so far away from it, It would be great if you can guide me a little more.

Bye there, and thanks again!

 

[micromass]

No, you haven't used the Lagrange multipliers wrong, your calculations are correct (except that you diveded by z somewhere, which could be 0). However, you searched conditions on $$\lambda$$. You should look for conditions on (x,y,z) instead.

So, to resume, you have following situation

$$\left\{\begin{array}{l}<br /> \lambda=yz\\<br /> \lambda=xy\\<br /> \lambda=xz\\<br /> x+y+z=w<br /> \end{array} \right.$$

So we got xy=yz=xz.
Now there are a few situations which can occur:

1) x=0
Then yz=0, so either y or z equals zero. The fourth equation then yields that (x,y,z) is either the point (0,0,w) or (0,w,0)

2) y=0
analogous to 1, this gives the point (w,0,0) and (0,0,w)

3) z=0
analogous to 1, this gives the point (w,0,0) and (0,w,0)

4) neither x,y or z equal 0
Then xy=yz=xz must imply that x=y=z. The fourth equation then yields that x=y=z=w. So this gives us the point (w,w,w).


So the only points which satisfy the 4 equations are (w,0,0), (0,w,0), (0,0,w) and (w,w,w). Now you can check manually which of these 4 maximizes f.

 

[Telemachus]

Thanks.