A unit sphere is on a level surface in the rain. What is the largest right rectangular prism, by volume, that can be placed under the sphere and not get rained on?

I’m enjoying this ‘sheltered’ series Jer. Where are you taking us next?

Assuming the best position for the sheltered block is with two of its base corners on the projected circle that defines the edge of the dry region...
Label the vertices above these corners A and B with M being the mid-point of AB. Let T be the mid-point of the other upper edge parallel to AB. The block and the sphere will touch at T.
Let O be the centre of the sphere and C be a point vertically below O on the horizontal line MT, so that C is the centre of a circle, radius 1, through A and B.

Define two variables:      z = |OC| and   x = |CM|, so that the block has
dimensions:

length = 2*sqrt(1 – x²),       height = 1 – z,          width = x – sqrt(1 – z²)

then Volume,     V = 2(1 – z)*sqrt(1 – x²)[(x – sqrt(1 – z²)]

For maximum V in  0 < x,z <1 we would expect partial derivatives as follows:

dV/dz = 0:         2(1 – x²)^0.5[(1 – z)z(1 – z²)^-0.5 - (x - (1 – z²)^0.5)] = 0

dV/dx = 0:        2(1 – z)[(x – (1 – z²)^0.5)(-x(1 – x²)^-0.5) + (1 - x²)^0.5] = 0

Knowing that the maximum cannot occur at x = 1 or z = 1, and after much rearrangement, these equations simplify to:

            x = (1 – z)(1 + 2z)/sqrt(1 – z²)                            (1)

and       1 – 2x² + x*sqrt(1 – z²) = 0                                (2)

Using (1) in (2), eventually gives:    z(6z² – z – 3) = 0   and, since we know
that z > 0, this gives:
                                    z = (1/12)(1 + sqrt(73))  = 0.7953336.. 

and then from (1):         x = (1/6)(sqrt(19 + sqrt(73))  = 0.8747063..

Hence the following dimensions for maximum volume:

length = (2/3)*sqrt(17 – sqrt(73))  = 0.9693065.. 

height = (1/12)*sqrt(11 – sqrt(73))  = 0.2046663..

width = (1/6)[sqrt(19 + sqrt(73)) – sqrt(35 – sqrt(73))/sqrt(2)]  = 0.2685344..

Maximum volume = (1/108)*sqrt(76126 – 8906*sqrt(73))  = 0.0532730..

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I’ve had to leave out much of the detailed working, but I hope it still makes some sense and doesn’t have too many errors.

Hopefully not abridged too far !

Posted by Harry on 2012-05-09 20:02:45