Final answer:
The dimensions of a rectangular box with the maximum volume and a fixed surface area of 64 square units would be those of a cube, considering a cube has the most efficient distribution of surface area. The side of this cube is the square root of the surface area divided by 6. The volume is then the cube of this side length.
Step-by-step explanation:
The question we are tackling involves finding the dimensions of a rectangular box with the maximum volume given a fixed surface area of 64 square units. This is a classic optimization problem in mathematics. To solve it, we'll use calculus and the method of Lagrange multipliers or geometric reasoning due to the symmetrical nature of the problem.
To begin with, let's denote the length, width, and height of the box as l, w, and h respectively. The surface area (SA) and volume (V) of a rectangular box are given by the formulas SA = 2lw + 2lh + 2wh and V = lwh. Our goal is to maximize V while keeping SA fixed at 64 square units.
The problem can be simplified by realizing that, for a given surface area, a cube will always have the maximum volume since the distribution of the area over the six faces is most efficient in that shape. Since we know the surface area of the cube (which is 6 times one of its face's area, since all faces are equal), we can calculate the length of a side of this cube by dividing the total surface area by 6 and then taking the square root of this result.
To calculate the maximum volume from this, we cubed the side length of the cube. With a total surface area of 64 square units, the side of the cube would be √(64/6), and the volume would be (√(64/6))^3 cubic units.