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Thread: 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:$\displaystyle 3(\underbrace{xy}_{\text{top}}+\underbrace{2xz+2yz }_{\text{sides}})+\underbrace{2xy}_{\text{bottom}}$


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

    subject to the constraint

    $\displaystyle xyz=2250$.....[2]

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

    $\displaystyle z=\frac{2250}{xy}$

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

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

    Differentiate [3] with respect to x and y:

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

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

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

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

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

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

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

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

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

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

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

    Therefore, we have a relative minimum.


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

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

    $\displaystyle g(x,y,z)=xyz$

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

    $\displaystyle {\nabla}g=yxi+xzj+xyk$

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

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

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

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

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

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

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

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

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