# Given a positive number decomposed on three summands find the maximum of its product

• Oct 30th 2010, 07:29 AM
Ulysses
Given a positive number decomposed on three summands find the maximum of its product
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.
• Oct 30th 2010, 09:44 AM
emakarov
This is an interesting problem because airlines, subways, etc. often give a restriction on luggage as the maximum on the sum of height, width and depth. The problem then is to devise a container with the maximum volume that conforms to the restriction.

Let z = w - x - y; substitute z into f(x,y,z) to get a function of x and y and a (fixed) parameter w. Then find its partial derivatives. Equating the derivatives to zero has only one solution where x and y are not zero.
• Oct 30th 2010, 11:16 AM
Soroban
Hello, Ulysses!

Quote:

Decompose a positive number on three non-negative summands
so that their product is maximum.

We have: . $N \:=\:x + y + z \quad\Rightarrow\quad z \:=\:N - x - y$ .[1]

We want to maximize: . $P \:=\:xyz$ .[2]

Substitute [1] into [2]: . $P \;=\;xy(N - x - y)$

. . . . . . . . . . . . . . . . . $P \;=\;Nxy - x^2y - xy^2$

Set the partial derivatives equal to zero and solve.

. . $\begin{array}{cccccccccc}
\dfrac{\partial P}{\partial x} &=& Ny - 2xy - y^2 &=& 0 & [3] \\ \\[-2mm]
\dfrac{\partial P}{\partial y} &=& Nx - x^2 - 2xy &=& 0 & [4] \end{array}$

From [3], we have: . $y(N-2x - y) \:=\:0$
From [4], we have: . $x(N - x - 2y) \:=\:0$

Since neither $\,x$ nor $\,y$ can be zero
. . (they would not produce maximum $\,P),$
we have this system of equations:

. . $\begin{array}{ccc}N - 2x - y &=& 0 \\ N - x - 2y &=& 0 \end{array}\quad\Rightarrow\quad \begin{array}{ccc}2x + y &=& N \\ x + 2y &=& N \end{array}$

Solve the system: . $x \:=\:\frac{N}{3},\;y \:=\:\frac{N}{3}$

Substitute into [1]: . $z \:=\:N - \frac{N}{3} - \frac{N}{3} \quad\Rightarrow\quad z \:=\:\frac{N}{3}$

For maximum product, the three summands must be equal.

• Oct 30th 2010, 12:25 PM
Ulysses
Thank you both. Great explanations, and great job.

Bye there!