We are tasked with determining the dimensions of a rectangular box with a square base and an open top that minimizes the cost of materials, subject to the constraint that the box must hold a volume of 400 cubic centimeters.

Step 1: Define the variables

Let:

• $x$ be the side length of the square base (in cm).
• $h$ be the height of the box (in cm).

Step 2: Volume constraint

The volume $V$ of the box is given by the formula: $V = x^2 \cdot h$ Since the volume is constrained to 400 cubic centimeters, we have: $x^2 \cdot h = 400$ $h = \frac{400}{x^2}$

Step 3: Cost of materials

• The cost of the base is $5$ pounds per square centimeter. The area of the base is $x^2$, so the cost for the base is: $\text{Cost of base} = 5 \cdot x^2$
• The cost of the sides is $2$ pounds per square centimeter. There are four sides, each with an area of $x \cdot h$, so the total cost of the sides is: $\text{Cost of sides} = 4 \cdot (x \cdot h) = 4 \cdot x \cdot h$ Using $h = \frac{400}{x^2}$, we substitute this into the expression for the cost of the sides: $\text{Cost of sides} = 4 \cdot x \cdot \frac{400}{x^2} = \frac{1600}{x}$

Step 4: Total cost function

The total cost $C(x)$ is the sum of the costs of the base and the sides: $C(x) = 5 \cdot x^2 + \frac{1600}{x}$

Step 5: Minimize the cost

To minimize the cost, we need to find the critical points by taking the derivative of $C(x)$ with respect to $x$ and setting it equal to zero. First, we compute the derivative: $C'(x) = 10x - \frac{1600}{x^2}$ Now set the derivative equal to zero and solve for $x$: $10x - \frac{1600}{x^2} = 0$ Multiply through by $x^2$ to eliminate the denominator: $10x^3 = 1600$ $x^3 = 160$ $x = \sqrt[3]{160}$ Approximating this value: $x \approx 5.43 \, \text{cm}$

Step 6: Find the height

Using $x \approx 5.43$ cm, we can find the height $h$ from the volume equation: $h = \frac{400}{x^2} = \frac{400}{(5.43)^2} \approx 13.57 \, \text{cm}$

Step 7: Verify the second derivative

To ensure that this critical point is a minimum, we check the second derivative of $C(x)$: $C''(x) = 10 + \frac{3200}{x^3}$ Since $C''(x) > 0$ for all $x > 0$, the function is concave up at $x \approx 5.43$, confirming that this critical point is a minimum.

Conclusion

The dimensions that minimize the cost are approximately:

• Side length of the square base: $x \approx 5.43$ cm
• Height of the box: $h \approx 13.57$ cm

Would you like further details on any of the steps?

Here are 5 related questions to explore this concept further:

1. What would happen to the optimal dimensions if the volume of the box increased to 600 cubic centimeters?
2. How would the optimal dimensions change if the cost of the base material doubled?
3. What is the impact of having a closed top on the cost and dimensions?
4. How can the cost function be minimized using a graphical approach?
5. What if the base is not square but rectangular—how would the optimization process differ?

Tip: Always check the second derivative when determining whether a critical point is a minimum or maximum.