A worker is to construct an open rectangular box with a square base and a volume of 192 $\mathrm{ft}^{3}$. If material for the bottom costs $\$ 6 / \mathrm{ft}^{2}$ and material for the sides costs $\$ 8 / \mathrm{ft}^{2}$, what dimensions will result in the least expensive box? What is the minimum cost?

Step 1 ：The problem is asking for the dimensions of the box that will minimize the cost of the materials. The cost of the materials is determined by the surface area of the box, since the cost per square foot is given. The volume of the box is also given, which provides a relationship between the dimensions of the box.

Step 2 ：The volume of a rectangular box with a square base and height h is given by \(s^2h = 192\), where s is the side length of the base. We can solve this equation for h to get \(h = 192/s^2\).

Step 3 ：The surface area of the box is given by \(A = s^2 + 4sh\), where A is the area, s is the side length of the base, and h is the height. Substituting the expression for h from the volume equation into the area equation gives \(A = s^2 + 4s(192/s^2) = s^2 + 768/s\).

Step 4 ：The cost of the materials is given by \(C = 6s^2 + 8(4sh) = 6s^2 + 32sh\). Substituting the expression for h from the volume equation into the cost equation gives \(C = 6s^2 + 32s(192/s^2) = 6s^2 + 6144/s\).

Step 5 ：To minimize the cost, we need to find the derivative of the cost function with respect to s and set it equal to zero. This will give us the value of s that minimizes the cost. We can then substitute this value of s back into the volume equation to find the corresponding value of h.

Step 6 ：The critical points are [8, -4 - 4*sqrt(3)*I, -4 + 4*sqrt(3)*I]. The minimum cost occurs when the side length of the base is 8 ft and the height is 3 ft.

Step 7 ：The minimum cost is \(\boxed{1152}\) dollars. The dimensions that will result in the least expensive box are a base with side length 8 ft and height 3 ft.