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Thread: Maximum Volume

  1. #1
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    Maximum Volume

    Find the maximum volume of a box with faces parallel to the
    xy, xz, yz-planes that lies entirely inside the region enclosed by the cones

    z = sqrt(x^2 + 2y^2) - 2 and

    z = 2 - sqrt(x^2 + 2y^2)

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  2. #2
    Super Member malaygoel's Avatar
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    The enclosed region is symmetric about xy-plane.

    So, if z=k is on of the plane, then the other is z=-k.

    If we cut the enclosed area by the plane z=k, we will get an ellipse
    $\displaystyle x^2+2y^2=(2-k)^2$

    or,$\displaystyle \frac{x^2}{(2-k)^2}+\frac{y^2}{0.5(2-k)^2}=1$

    The area of the rectangle having largest area(with sides parallel to ellipse axes) is $\displaystyle \sqrt 2 (2-k)^2$

    Hence, the volume of box is $\displaystyle 2\sqrt 2 k(2-k)^2$

    For maximum volume, value of k is found to be $\displaystyle k=\frac{2}{3}$
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  3. #3
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    What method are you supposed to use ?
    It looks like a Lagrange Multiplier problem to me.
    Choose a point say $\displaystyle (p,q,r)$ on the cone
    $\displaystyle z = 2 - \sqrt(x^2 + 2y^2)$
    in the first octant, (so $\displaystyle p,q,r$ are all positive ), and form your box based on this point. It will have a volume of $\displaystyle 8pqr$.
    Now let
    $\displaystyle W = 8pqr - \lambda\{2 - r - \sqrt(p^2 + 2q^2)\}$
    and differentiate partially wrt $\displaystyle p,q,r$ and $\displaystyle \lambda$.
    Put each partial derivative equal to zero and solve the resulting equations simultaneously.
    I get the values $\displaystyle p = 2\sqrt(2)/3, q = r = 2/3$,
    in which case the maximum volume is $\displaystyle 64\sqrt(2)/27.$
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