To find the dimensions of the rectangular box that will minimize the cost, we should first express the cost as a function of the dimensions of the box and then find its minimum value.

1. Define variables:

   • Let $x$ be the length of each side of the square base (and top) of the box, in feet.
   • Let $h$ be the height of the box, in feet.

2. Volume constraint: The volume of the box is given by: $V = x^2 \cdot h = 960 \, \text{ft}^3.$ From this, we can express the height $h$ in terms of $x$: $h = \frac{960}{x^2}.$

3. Cost function:

   • The area of the bottom (base) is $x^2$.
   • The area of the top is also $x^2$.
   • The area of the four sides is $4 \cdot (x \cdot h) = 4xh = \frac{3840}{x}$ (since $4x \cdot \frac{960}{x^2} = \frac{3840}{x}$).

   The cost function $C(x)$ in dollars can be written as:

   • Cost of the bottom: $0.15 \times x^2$.
   • Cost of the top: $0.10 \times x^2$.
   • Cost of the sides: $0.015 \times \frac{3840}{x}$.

   Therefore, the total cost is: $C(x) = 0.15x^2 + 0.10x^2 + 0.015 \cdot \frac{3840}{x}.$ Simplified: $C(x) = 0.25x^2 + \frac{57.6}{x}.$

4. Minimize the cost: To find the minimum cost, take the derivative of $C(x)$ with respect to $x$ and set it equal to zero: $C'(x) = 0.50x - \frac{57.6}{x^2}.$ Set $C'(x) = 0$: $0.50x = \frac{57.6}{x^2}.$ $0.50x^3 = 57.6.$ $x^3 = \frac{57.6}{0.50}.$ $x^3 = 115.2.$ $x = \sqrt[3]{115.2}.$

   Using a calculator: $x \approx 4.88 \, \text{ft}.$

5. Find $h$: Using the volume constraint: $h = \frac{960}{x^2}.$

   Substituting $x \approx 4.88$: $h \approx \frac{960}{(4.88)^2}.$

   Calculating: $h \approx \frac{960}{23.8144} \approx 40.31 \, \text{ft}.$

6. Conclusion: The dimensions that minimize the cost of the box are approximately $x \approx 4.88 \, \text{ft}$ for the side of the square base and top, and $h \approx 40.31 \, \text{ft}$ for the height.