A cardboard box with an open top and a square bottom is to have a volume of 45 ft3 . Use a table utility to determine the dimensions of the box to the nearest 0.1 foot that will minimize the amount of cardboard used to construct the box.

Let the side of the bottom square be x, and the height of the box, h.

The volume is x²h=45
or
h=45/x²

The total surface area (5 faces) is
S=x²+4(x*h)=x(x+4h)
Make a table of 3 columns, comprising x, h and S. The value of x could vary between 4 and 5 in 0.1 ft intervals.

To determine the dimensions of the box that will minimize the amount of cardboard used, we can use the concept of optimization.

Let's start by assigning variables to the dimensions of the box. Let's say the length and width of the square bottom are both x, and the height of the box is h.

Given that the volume of the box is 45 ft^3, we can write the equation:

V = x^2 * h = 45

To minimize the amount of cardboard used, we need to minimize the surface area of the box. The surface area consists of the area of the bottom and the four sides.

The area of the bottom is easily calculated as A_bottom = x^2.

The area of the four sides can be calculated by multiplying the height by the perimeter of the square bottom. The perimeter of a square is given by P = 4x, so the area of the four sides is A_sides = 4xh.

The total surface area of the box is then given by A_total = A_bottom + A_sides = x^2 + 4xh.

To minimize the surface area, we can take the derivative of A_total with respect to x and equate it to zero:

dA_total/dx = 2x + 4h = 0

Solving this equation for h gives h = -x/2.

Now we can substitute the value of h back into the volume equation:

x^2 * (-x/2) = 45

Simplifying:

x^3 = -90

Now, we can use a table utility or a graphing calculator to find the value of x that satisfies this equation. By trying different values of x and solving for h, we can calculate the corresponding surface area A_total. The dimension set that results in the smallest A_total will be the dimensions that minimize the amount of cardboard used to construct the box.