To solve this problem step by step:

Step a) Formula for Volume

The dimensions are:

• Width of base: $x$
• Length of base: $2x$
• Height: $y$

The volume of the container (no cover) is given as: $\text{Volume} = \text{Base Area} \times \text{Height}$ $\text{Volume} = (x \cdot 2x) \cdot y = 2x^2y$

Since the volume is fixed at $12 \, \text{m}^3$, we have: $2x^2y = 12$ $y = \frac{12}{2x^2} = \frac{6}{x^2}$

Step b) Formula for Cost

The cost of the materials is the sum of the cost of the base and the cost of the walls:

1. Base cost: $40 \cdot \text{Area of the base} = 40 \cdot (x \cdot 2x) = 40 \cdot 2x^2 = 80x^2$
2. Wall costs: There are 4 walls:
   • Two walls of area $x \cdot y$: Total cost = $2 \cdot (20 \cdot xy) = 40xy$
   • Two walls of area $2x \cdot y$: Total cost = $2 \cdot (20 \cdot 2xy) = 80xy$
   • Total wall cost = $40xy + 80xy = 120xy$

The total cost is: $\text{Cost} = \text{Base cost} + \text{Wall cost}$ $\text{Cost} = 80x^2 + 120xy$

Substitute $y = \frac{6}{x^2}$ into the cost formula: $\text{Cost} = 80x^2 + 120x \left(\frac{6}{x^2}\right)$ $\text{Cost} = 80x^2 + \frac{720}{x}$

Step c) Minimizing the Cost

To minimize the cost, differentiate $\text{Cost}$ with respect to $x$, set the derivative to zero, and solve for $x$: $\frac{d}{dx} \left( 80x^2 + \frac{720}{x} \right) = 160x - \frac{720}{x^2}$ Set $\frac{d}{dx} = 0$: $160x - \frac{720}{x^2} = 0$ $160x = \frac{720}{x^2}$ $160x^3 = 720$ $x^3 = \frac{720}{160} = 4.5$ $x = \sqrt[3]{4.5} \approx 1.65 \, \text{m}$

Now substitute $x \approx 1.65$ into $y = \frac{6}{x^2}$ to find $y$: $y = \frac{6}{(1.65)^2} \approx \frac{6}{2.7225} \approx 2.20 \, \text{m}$

Dimensions:

• Width of the base: $x \approx 1.65 \, \text{m}$
• Length of the base: $2x \approx 2 \cdot 1.65 = 3.30 \, \text{m}$
• Height: $y \approx 2.20 \, \text{m}$

Final Answer:

• Height of the container: $2.20 \, \text{m}$
• Width of the base: $1.65 \, \text{m}$
• Length of the base: $3.30 \, \text{m}$

If you have further questions or need clarification, let me know!

Five Related Questions:

1. How do we calculate surface area of a container with different shapes?
2. What if the cost of the materials changes? How does it affect optimization?
3. How does minimizing cost change if the container has a lid?
4. Can we generalize this problem for different volume constraints?
5. What happens to the dimensions if the width-to-length ratio changes?

Tip: Always verify the derivative calculations when minimizing costs or maximizing efficiency.