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Math Help - Given a positive number decomposed on three summands find the maximum of its product

  1. #1
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    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.
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  2. #2
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    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.
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  3. #3
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    Hello, Ulysses!

    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}<br />
\dfrac{\partial P}{\partial x} &=& Ny - 2xy - y^2 &=& 0 & [3] \\ \\[-2mm]<br />
\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.

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  4. #4
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    Thank you both. Great explanations, and great job.

    Bye there!
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