To minimize the cost of constructing this rectangular box with a square base and top, we need to find the dimensions that will minimize the surface area cost, given that the volume is fixed at 931 cubic feet.

Let:

• $x$ = length of one side of the square base (in feet),
• $h$ = height of the box (in feet).

Step 1: Set up Volume Equation

Since the volume $V$ is 931 cubic feet: $V = x^2 \cdot h = 931$ This equation allows us to express $h$ in terms of $x$: $h = \frac{931}{x^2}$

Step 2: Set up Surface Area and Cost Equation

The surface area $S$ consists of:

• Bottom area: $x^2$ (square base)
• Top area: $x^2$ (square top)
• Four sides, each with area $x \cdot h$

The cost for each part is:

• Bottom: $x^2 \cdot 0.15$ (15 cents per square foot)
• Top: $x^2 \cdot 0.10$ (10 cents per square foot)
• Sides: $4 \cdot (x \cdot h) \cdot 0.015$ (1.5 cents per square foot for four sides)

The total cost $C$ is therefore: $C = 0.15x^2 + 0.10x^2 + 4 \cdot 0.015 \cdot (x \cdot h)$ Substitute $h = \frac{931}{x^2}$: $C = 0.25x^2 + 0.06x \cdot \frac{931}{x^2}$ Simplify: $C = 0.25x^2 + \frac{55.86}{x}$

Step 3: Minimize the Cost Function

To find the minimum cost, take the derivative of $C$ with respect to $x$ and set it to zero: $\frac{dC}{dx} = 0.5x - \frac{55.86}{x^2} = 0$ Multiply by $x^2$ to clear the fraction: $0.5x^3 = 55.86$ Solving for $x$: $x^3 = \frac{55.86}{0.5} = 111.72$ $x = \sqrt[3]{111.72} \approx 4.8 \text{ feet}$

Step 4: Find $h$

Using $h = \frac{931}{x^2}$ with $x \approx 4.8$: $h = \frac{931}{(4.8)^2} \approx \frac{931}{23.04} \approx 40.4 \text{ feet}$

Solution

The dimensions that minimize the cost are approximately:

• Length of one side of the base: $x \approx 4.8$ feet
• Height of the box: $h \approx 40.4$ feet

Would you like additional details on any steps?

Related Questions:

1. How would changing the cost of the sides impact the optimal dimensions?
2. What is the cost if the height is increased by 10%?
3. How does fixing the base to a particular dimension affect the height?
4. Could we minimize surface area instead of cost? How would that change the solution?
5. What are the dimensions if the cost for the bottom and top are equal?

Tip:

When optimizing dimensions with constraints like volume, express as many variables in terms of one and substitute into cost or surface area functions. This simplifies finding the extremum.