What whole number dimensions would allow the volume to be maximized while keeping the surface area at most 160 square feet?

Answer 1

Volume is maximized for a fixed surface area with a sphere, with radius = 3 feet, giving volume = 113 feet³

However, the rounding down to a whole number causes a paradox, making a cube with more volume.

Volume of a sphere is (4/3)pi(r)³

^{Surface area of a sphere = 4pi(r)2 < 160}

r² < 160/4pi = 40/pi

r < sqr40/pi = 3.57 = 3 feet rounded to greatest whole number less than actual.

that dimension for radius r will maximize volume of a sphere. A sphere provides the maximum volume for a given surface area

V < 4/3 pi (2sqr10/pi)³ = 4/3pi(3)³ = 113 cubic feet = max volume given surface area < 160 feet

rounded down to nearest integer

If, by chance, you meant a cube, then it's surface area = x³ = volume

and surface area 160 = 6x² = 6 sides each with area=x²

6x² = 160

x² = 160/6 = 80/3

x= sqr(80/3) = 5.16

dimensions of the cube are 5x5x5 feet, for whole numbers

making volume 125 cubic feet

That's 125 > 113 for the sphere, because of the rounding down to an integer.

we lost 0.57 for the radius by rounding down for the sphere but only lost 0.16 for the side of a cube

So, if those are the constraints, an integer dimension, the cube will have greater volume than the sphere, which is paradoxical, but only due to the rounding down.