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Thread: Finding the dimensions of a rectangular container?

  1. #1
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    Finding the dimensions of a rectangular container?

    What is the largest volume if the surface area of the container is 96 m^2?

    My work:
    V = lwh
    V = xyz

    SA = 96 = xy + 2xz 2yz
    z = (96 - xy)/(2x + 2y)

    V = xy(96 - xy)/(2x + 2y)
    V = (96xy - (x^2)(y^2))/(2x + 2y)

    Not sure if my work is on track... But can somebody show me how to get the answer? It's x = y = z = 4m.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by AlphaRock View Post
    What is the largest volume if the surface area of the container is 96 m^2?

    My work:
    V = lwh
    V = xyz

    SA = 96 = xy + 2xz 2yz
    z = (96 - xy)/(2x + 2y)

    V = xy(96 - xy)/(2x + 2y)
    V = (96xy - (x^2)(y^2))/(2x + 2y)

    Not sure if my work is on track... But can somebody show me how to get the answer? It's x = y = z = 4m.
    From what you have in your SA equation, the container has no lid! If it has a lid, then $\displaystyle SA=96=2xy+2xz+2yz\implies z=\frac{48-xy}{x+y}$

    Therefore, $\displaystyle V(x,y)=\frac{48xy-x^2y^2}{x+y}$.

    Now you have to find the critical point of this multivariable function.

    Therefore, $\displaystyle \frac{\partial V}{\partial x}=\frac{(x+y)(48y-2xy^2)-(48xy-x^2y^2)}{(x+y)^2}=\frac{48y^2-x^2y^2-2xy^3}{(x+y)^2}$

    and

    $\displaystyle \frac{\partial V}{\partial y}=\frac{(x+y)(48x-2x^2y)-(48xy-x^2y^2)}{(x+y)^2}=\frac{48x^2-x^2y^2-2x^3y}{(x+y)^2}$

    Now, set $\displaystyle \frac{\partial V}{\partial x}=0$ and $\displaystyle \frac{\partial v}{\partial y}=0$. Observe that

    $\displaystyle \frac{\partial V}{\partial x}=0\implies 48y^2-x^2y^2-2xy^3=0\implies x^2y^2=48y^2-2xy^3$

    Therefore, $\displaystyle \frac{\partial V}{\partial y}=0\implies 48x^2-48y^2+2xy^3-2x^3y=0$

    But this implies that $\displaystyle 48(x^2-y^2)-2xy(x^2-y^2)=(48-2xy)(x^2-y^2)=0$. Therefore $\displaystyle xy=24$ or $\displaystyle x=y$.

    If $\displaystyle x=y$ then $\displaystyle \frac{\partial V}{\partial x}=0\implies 48x^2-x^4-2x^4=0\implies$ $\displaystyle 16x^2-x^4=0\implies x^2(16-x^2)=0$

    Therefore, $\displaystyle x^2=0\implies x=0$ or $\displaystyle 16-x^2=0\implies x=\pm 4$. Since length can't be negative, we have $\displaystyle x=4$.

    As a result, $\displaystyle y=4$. Therefore, $\displaystyle z=\frac{48-(4)(4)}{4+4}=\frac{32}{8}=4$.

    Therefore, the rectangular box that has maximum volume and SA = 96 sq. meters is a cube with dimensions $\displaystyle x=y=z=4$ meters.

    You can verify that this is a maximum by applying the second derivative test at the point $\displaystyle (4,4,4)$.

    Does this make sense?

    (I'm sure you could have applied Lagrange Multipliers here to $\displaystyle f(x,y,z)=xyz$ with the restriction $\displaystyle 2xy+2xz+2yz=96$ (which could cut back on the calculations that are required), but this is what came to my mind first.)
    Last edited by Chris L T521; Apr 3rd 2010 at 02:24 PM. Reason: fixed minor typo
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