To solve this problem, let’s break it into steps:

Problem Setup

1. Dimensions of the container:

   • Let $w$ be the width of the base.
   • The length of the base is $2w$, since the length is twice the width.
   • Let $h$ be the height of the container.

2. Volume constraint: $\text{Volume} = \text{Base area} \times \text{Height} = (w \cdot 2w) \cdot h = 2w^2h.$ Since the volume is given as 12 cubic meters, we have: $2w^2h = 12 \quad \Rightarrow \quad h = \frac{12}{2w^2} = \frac{6}{w^2}.$

3. Cost of materials:

   • The base has an area of $w \cdot 2w = 2w^2$, and the material for the base costs $14 per square meter.
   • The four sides include two rectangles with area $w \cdot h$ and two rectangles with area $2w \cdot h$. Their total area is: $2(w \cdot h) + 2(2w \cdot h) = 6wh.$ Material for the sides costs $9 per square meter.

   The total cost is: $C = 14(2w^2) + 9(6wh).$

4. Substitute $h = \frac{6}{w^2}$ into the cost function: Substituting for $h$ into $C$: $C = 14(2w^2) + 9(6w \cdot \frac{6}{w^2}),$ which simplifies to: $C = 28w^2 + 54 \cdot \frac{1}{w}.$

Optimization

To find the minimum cost, we differentiate $C$ with respect to $w$ and set the derivative to zero: $\frac{dC}{dw} = 56w - 54 \cdot \frac{1}{w^2}.$ Setting $\frac{dC}{dw} = 0$: $56w = 54 \cdot \frac{1}{w^2} \quad \Rightarrow \quad 56w^3 = 54 \quad \Rightarrow \quad w^3 = \frac{54}{56} = \frac{27}{28}.$ Thus: $w = \left(\frac{27}{28}\right)^{\frac{1}{3}}.$

Calculate Cost

Using $w \approx \left(\frac{27}{28}\right)^{\frac{1}{3}}$, substitute back into the cost function $C$ to compute the minimum cost. Let me calculate this final step.# Given values for the calculation from sympy import symbols, solve, diff

Define the variables

w = symbols('w', positive=True) cost_function = 28 * w**2 + 54 / w

Differentiate the cost function

dc_dw = diff(cost_function, w)

Solve for w

optimal_w = solve(dc_dw, w)[0]

Compute h and the total cost

h = 6 / (optimal_w**2) # h as a function of optimal w min_cost = cost_function.subs(w, optimal_w)

optimal_w_value = float(optimal_w.evalf()) h_value = float(h.evalf()) min_cost_value = float(min_cost.evalf())

optimal_w_value, h_value, min_cost_value(0.9879506340125243, 6.147248389411262, 81.98790224064288)(0.9879506340125243, 6.147248389411262, 81.98790224064288)successThe optimized dimensions and minimum cost for the container are:

• Width ($w$): Approximately $0.988$ meters.
• Height ($h$): Approximately $6.147$ meters.
• Minimum cost: Approximately $81.99.

Let me know if you'd like a detailed explanation of the calculations or further questions on this! Here are some related questions:

1. What is the area of the base and sides at the minimum cost?
2. How would the cost change if the material prices increased by 10%?
3. What happens if the volume constraint is increased to 20 cubic meters?
4. Could the cost function have multiple minima? Why or why not?
5. How would the solution differ if the container had a closed top?

Tip: When optimizing, always verify critical points with the second derivative or boundary conditions to ensure a global minimum.