Final answer:
To find the dimensions of the box for which the volume is as large as possible, we need to maximize the volume while keeping the surface area constant. By solving for the dimensions using the given information and trial and error, we find that the dimensions that yield the maximum volume are: Length (l) = 2 feet, Width (w) = 4 feet, and Height (h) = 4 feet.
Step-by-step explanation:
To find the dimensions of the box for which the volume is as large as possible, we need to maximize the volume while keeping the surface area constant. The surface area of an open rectangular box is given by the formula:
Surface Area = 2lw + 2lh + 2wh
Given that the length (l) is 2 feet and the surface area is 32 square feet, we can substitute these values into the formula and solve for the other dimensions:
32 = 2(2)(w) + 2(2)(h) + 2(w)(h)
16 = 2w + 2h + wh
Simplifying the equation further:
8 = w + h + 0.5wh
To find the maximum volume, we can use calculus to take the derivative of the volume function (V = lwh) with respect to one of the variables and set it equal to zero. However, since this is a high school level question, we can use trial and error to find the dimensions that yield the maximum volume.
By trying different values for the width and height, we can find that the dimensions for which the volume is as large as possible are:
Width (w) = 4 feet
Height (h) = 4 feet
Therefore, the dimensions of the box that maximize the volume are: Length (l) = 2 feet, Width (w) = 4 feet, and Height (h) = 4 feet.