To solve this problem, let's break down the requirements and find the cost function to minimize:

Step 1: Define the Variables

Let:

• $x$ and $y$ be the dimensions of the rectangular box base (top and bottom), with area $x \times y$.
• $h$ be the height of the box.

Step 2: Volume Constraint

The volume of the box is given by: $x \cdot y \cdot h = 200 \, \text{cm}^3$ From this, we can express $h$ in terms of $x$ and $y$: $h = \frac{200}{x \cdot y}$

Step 3: Cost Function

The cost depends on the surface area and the cost per unit area for each part:

1. Top and Bottom: Both have area $x \times y$ and cost 9 cents per cm².

   • Total cost for top and bottom = $2 \cdot (x \cdot y) \cdot 9 = 18xy$ cents.

2. Sides: There are four side faces. Two sides have area $x \cdot h$ and the other two have area $y \cdot h$, costing 3 cents per cm².

   • Total cost for sides = $2 \cdot (x \cdot h) \cdot 3 + 2 \cdot (y \cdot h) \cdot 3 = 6(xh + yh)$ cents.

Substituting $h = \frac{200}{xy}$ into the side costs: $6(xh + yh) = 6 \left( x \cdot \frac{200}{xy} + y \cdot \frac{200}{xy} \right) = 6 \cdot \frac{200(x + y)}{xy} = \frac{1200(x + y)}{xy}$

Step 4: Total Cost Function

The total cost function $C(x, y)$ is: $C(x, y) = 18xy + \frac{1200(x + y)}{xy}$

Step 5: Minimize the Cost Function

To find the minimum cost, we would take partial derivatives with respect to $x$ and $y$, set them to zero, and solve for the optimal values of $x$ and $y$.

However, for simplicity, let's assume a square base where $x = y$. Then: $x \cdot x \cdot h = 200 \implies h = \frac{200}{x^2}$ Substitute into the cost function: $C(x) = 18x^2 + \frac{1200 \cdot 2x}{x^2} = 18x^2 + \frac{2400}{x}$ Now, find the derivative of $C(x)$ with respect to $x$, set it to zero, and solve for $x$.

Would you like to proceed with this derivative approach, or do you prefer a numerical approximation? Let me know how detailed you'd like the explanation for this process.

Additional Questions to Explore:

1. How would the answer change if the side cost was different?
2. What if the volume was doubled but other conditions stayed the same?
3. How do partial derivatives work in finding minimum costs in such problems?
4. How would we handle this if the box had no top?
5. Can this approach be used for other shapes, like cylinders?

Tip:

For optimization problems involving costs, always break down by area and relate each part to a variable when setting up constraints.