Strategy:
You will need to make use of the formulas for volume and area of the shapes you have chosen.
... volume of a sphere = (4/3)πr³
... area of a sphere = 4πr²
You will have to use the volume equation to find the radius of the sphere with the desired volume. Then use the area formula to verify the area is smaller than for your rectangular prism. (A sphere has the smallest surface area for a given volume of any figure.)
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For your pyramid, it's a bit more complicated. You can use the formula for volume to find a relationship between the side length and the height. Then you can solve for height and put that expression into the equation for surface area. Now, you have an equation for surface area as a function of side length, which you can use to choose side lengths that give surface area in the range you want. You may find a graphing utility to be helpful for this.
... volume of a pyramid = (1/3)b^2h . . . . . b is the edge length of the base; h is the verical height.
... area of a pyramid = b^2 + 2bs, where s is the slant height: s=√((b/2)^ +h^2)
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The cylinder is solved essentially the same way the pyramid is solved, except the formulas are slightly different.
... volume of a cylinder = πr²h
... area of a cylinder = 2πr² + 2πrh
Numbers:
If I did it right, a pyramid with a base edge length of 4 or 5 inches (or somewhere between) should give a smaller surface area for the volume you want.
For the cylinder, the radius may be around 2 or 2.1 inches.
be an integer. Suppose there is a triangle with legs of length 16 and 
. Then by the Pythagorean theorem, the length of the hypotenuse should be
and 
are integers, then so are 
, 
, and 
.
is not a perfect square trinomial, which means for any integer