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Math Help - Question about maximizing area.

  1. #1
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    Question about maximizing area.

    A closed rectangle container with a square base must have a volume of 2250m^3. The material for the top and the sides of the container will cost $3/m^2. The material for teh bottom of the container will cost $2/m^2. Determine the dimensions of the box which will minimize the cost of the material for the box.

    Now I know to find mins you take the 1st deriv test, but how do I apply this to this question?
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  2. #2
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    Surface area is: 3(\underbrace{xy}_{\text{top}}+\underbrace{2xz+2yz  }_{\text{sides}})+\underbrace{2xy}_{\text{bottom}}


    Therefore, we want S=5xy+6xz+6yz........[1]

    subject to the constraint

    xyz=2250.....[2]

    Solve [1] for z and sub into [2]:

    z=\frac{2250}{xy}

    S=5xy+6x(\frac{2250}{xy})+6y(\frac{2250}{xy})

    =5xy+\frac{13500}{y}+\frac{13500}{x}....[3]

    Differentiate [3] with respect to x and y:

    S_{x}=5y-\frac{13500}{x^{2}}=0...[4]

    S_{y}=5x-\frac{13500}{y^{2}}=0...[5]

    Solve [4] for y and sub into [5]:

    y=\frac{2700}{x^{2}}

    5x-\frac{13500}{(\frac{2700}{x^{2}})^{2}}=5x-\frac{x^{4}}{540}=0

    Solving this for x, we find x=3\cdot{10^{\frac{2}{3}}}\approx{13.92}

    Then y=3\cdot{10^{\frac{2}{3}}}\approx{13.92}

    and z=\frac{5\cdot{10^{\frac{2}{3}}}}{2}\approx{11.60}

    Check to see if it is, indeed a minimum. Use the Second Partials Test.

    D=S_{xx}(x,y)S_{yy}(x,y)-S_{xy}^{2}(x,y)=75>0

    Also, S_{xx}(x,y)=10>0

    Therefore, we have a relative minimum.


    Now, LaGrange multipliers. See if we get the same thing.

    Let f(x,y,z)=5xy+6xz+6yz

    g(x,y,z)=xyz

    {\nabla}f=(5y+6z)i+(5x+6z)j+(6x+6y)k

    {\nabla}g=yxi+xzj+xyk

    (5y+6z)i+(5x+6z)j+(6x+6y)k={\lambda}(yzi+xzj+xyk)

    5y+6z=yz{\lambda}\rightarrow{\lambda}=\frac{5y+6z}  {yz}...[1]
    5x+6z=xz{\lambda}\rightarrow{\lambda}=\frac{5x+6z}  {xz}...[2]
    6x+6y=xy{\lambda}\rightarrow{\lambda}=\frac{6x+6y}  {xy}...[3]

    From [1] and [2], we get x=y

    From [2] and [3], we get z=\frac{5y}{6}

    Sub these into the constraint, xyz, and we get:

    \frac{5y^{3}}{6}=2250

    Solving for y, we get x=y=3\cdot{10^{\frac{2}{3}}}

    and z=\frac{5\cdot{10^{\frac{2}{3}}}}{2}

    Same as before. Good. It checks.
    Last edited by galactus; December 21st 2006 at 06:59 PM.
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