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  • Re: Maximize Cistern Volume -- (cut out 4 squares (at Corners) and disc

    From Richard Tobin@21:1/5 to [email protected] on Sat May 3 19:01:15 2025
    XPost: sci.lang, sci.math

    In article <[email protected]>,
    HenHanna <[email protected]> wrote:

    I let the derivative be 0 and solve it , and i get x = 1/2, 1/6

    at x=0 the slope is 1
    whereas at x=1/2, the slope is Zero!!!

    _______________

    at x=1/2, the slope is Zero!!!

    It's not obvious why, Can someone explain this?

    When x is 1/2 the side of the cistern has shrunk to zero, the height
    is 1/2, and the volume is zero. Physically, x can't exceed 1/2, but
    the formula just produces a negative length for the side of the
    cistern (along with a height greater then 1/2). That gives a positive
    volume (the negative length is squared), so x=1/2 is a minimum for the
    volume.

    -- Richard

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  • From efji@21:1/5 to All on Sun May 4 18:50:38 2025
    XPost: sci.lang, sci.math

    Le 04/05/2025 à 17:40, Ross Finlayson a écrit :
    It's sort of like soup cans vis-a-vis fish tins, or the idea
    that cans, to get the most volume of a cylinder, have
    at least two solutions.

    Nope.
    Classical optimization homework : "maximize the volume of a cylindric
    can under the constraint of a given surface", or its dual "minimize the
    surface of a cylindric can under the constraint of a given volume". Both problems have a unique solution : the height H is twice the radius R of
    the circular basis.

    --
    F.J.

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  • From HenHanna@21:1/5 to Richard Tobin on Tue May 6 19:35:47 2025
    XPost: sci.lang, sci.math

    On Sat, 3 May 2025 19:01:15 +0000, Richard Tobin wrote:

    In article <[email protected]>,
    HenHanna <[email protected]> wrote:

    I let the derivative be 0 and solve it , and i get x = 1/2, 1/6

    at x=0 the slope is 1
    whereas at x=1/2, the slope is Zero!!!

    _______________

    at x=1/2, the slope is Zero!!!

    It's not obvious why, Can someone explain this?

    When x is 1/2 the side of the cistern has shrunk to zero, the height
    is 1/2, and the volume is zero. Physically, x can't exceed 1/2, but
    the formula just produces a negative length for the side of the
    cistern (along with a height greater then 1/2). That gives a positive
    volume (the negative length is squared), so x=1/2 is a minimum for the volume.

    -- Richard

    ______________________________________

    Thank you... When i saw that this curve looks like the typical
    curve
    for
    y= x^3 + bx^2 + cx + d

    it made more sense that this "car" starts (at x=0) at the Top speed (of
    1)
    but gradually slows down to a halt (at x=1/2)


    What's not at all obvious (intuitive) for me is.... why or how
    the max Volume is achieved at x=1/6

    Could a little child guess that correctly ?

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  • From Richard Tobin@21:1/5 to [email protected] on Tue May 6 19:47:37 2025
    XPost: sci.lang, sci.math

    In article <[email protected]>,
    HenHanna <[email protected]> wrote:

    What's not at all obvious (intuitive) for me is.... why or how
    the max Volume is achieved at x=1/6

    Note that x=1/6 makes the total area of the sides equal to the area of
    the base (4/9). I wouldn't be surprised if that is a special case of
    some more general result.

    -- Richard

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  • From Mike Terry@21:1/5 to Richard Tobin on Tue May 6 23:29:51 2025
    XPost: sci.math

    On 06/05/2025 20:47, Richard Tobin wrote:
    In article <[email protected]>,
    HenHanna <[email protected]> wrote:

    What's not at all obvious (intuitive) for me is.... why or how
    the max Volume is achieved at x=1/6

    Note that x=1/6 makes the total area of the sides equal to the area of
    the base (4/9). I wouldn't be surprised if that is a special case of
    some more general result.

    -- Richard


    That makes sense - when we make the 4 cutout squares bigger, increasing their side length by a very
    small amount s, the effect on the cistern is, broadly
    a) increase the height by s, which /increases/ its volume by A.s
    [A being the area of the base]
    b) decrease the "radius" of the box by s, which /decreases/ its volume by B.s
    [B being the area of the sides]
    So at the maximum volume these two effects must cancel out, and we will have A = B. Yes there are
    higher order changes in volume involving s^2 and higher powers, but we neglect those as small
    compared to first order changes.

    This is in effect doing calculus from scratch, ignoring higher order terms in s to get the
    derivative dV/dx which is set to zero. The ignoring of higher terms is like what happens in the
    proof of the product rule for derivatives.


    Mike.

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