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

• Oct 30th 2010, 06:29 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
$\displaystyle w=x+y+z$, $\displaystyle w>0, x \geq{}0 , y \geq{}0 , z \geq{}0$
$\displaystyle f(x,y,z)=xyz$
$\displaystyle f_x=yz,f_y=xz,f_z=xy$
The thing is I don't see any maximum here, clearly I'm not setting the things right.
• Oct 30th 2010, 08:44 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, 10:16 AM
Soroban
Hello, Ulysses!

Quote:

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

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

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

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

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

Set the partial derivatives equal to zero and solve.

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

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

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

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

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

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

For maximum product, the three summands must be equal.

• Oct 30th 2010, 11:25 AM
Ulysses
Thank you both. Great explanations, and great job.

Bye there!