Find the minimum cost of a rectangular box of volume 110 cm3 whose top and bottom cost 8 cents per cm2 and whose sides cost 8 cents per cm2.

To find the minimum cost of a rectangular box with a given volume and specific costs for its surfaces, we can set up the problem using calculus and optimization techniques. Here's a step-by-step solution:

1. **Define the Variables:**
   - Let \( x \) be the length of the box (in cm).
   - Let \( y \) be the width of the box (in cm).
   - Let \( z \) be the height of the box (in cm).

2. **Volume Constraint:**
   The volume of the box is given by:
   \[
   V = x \cdot y \cdot z = 110 \, \text{cm}^3
   \]
   Therefore:
   \[
   z = \frac{110}{x \cdot y}
   \]

3. **Surface Area and Cost:**
   The cost of the top and bottom surfaces, which are both \( x \cdot y \), is \( 8 \) cents per cm\(^2\):
   \[
   \text{Cost of top and bottom} = 2 \cdot 8 \cdot (x \cdot y) = 16xy
   \]
   The cost of the four sides, each being \( xz \) and \( yz \), is also \( 8 \) cents per cm\(^2\):
   \[
   \text{Cost of the sides} = 8 \cdot (2 \cdot xz + 2 \cdot yz) = 16(xz + yz)
   \]

4. **Total Cost:**
   The total cost function \( C \) is:
   \[
   C = 16xy + 16(xz + yz)
   \]
   Substituting \( z = \frac{110}{x \cdot y} \) into the cost function:
   \[
   C = 16xy + 16 \left( x \cdot \frac{110}{xy} + y \cdot \frac{110}{xy} \right)
   \]
   Simplifying further:
   \[
   C = 16xy + 16 \left( \frac{110}{y} + \frac{110}{x} \right)
   \]
   \[
   C = 16xy + \frac{1760}{y} + \frac{1760}{x}
   \]

5. **Optimization:**
   To find the minimum cost, we need to take the partial derivatives of \( C \) with respect to \( x \) and \( y \) and set them to zero:
   \[
   \frac{\partial C}{\partial x} = 16y - \frac{1760}{x^2} = 0
   \]
   \[
   \frac{\partial C}{\partial y} = 16x - \frac{1760}{y^2} = 0
   \]

6. **Solve for \( x \) and \( y \):**
   From \(\frac{\partial C}{\partial x} = 0\):
   \[
   16y = \frac{1760}{x^2}
   \]
   \[
   y = \frac{110}{x^2}
   \]
   From \(\frac{\partial C}{\partial y} = 0\):
   \[
   16x = \frac{1760}{y^2}
   \]
   \[
   x = \frac{110}{y^2}
   \]

   Substituting \( y = \frac{110}{x^2} \) into \( x = \frac{110}{y^2} \):
   \[
   x = \frac{110}{\left(\frac{110}{x^2}\right)^2}
   \]
   \[
   x = \frac{110}{\frac{12100}{x^4}}
   \]
   \[
   x^5 = 12100
   \]
   \[
   x = \sqrt[5]{12100}
   \]

   Similarly, solving for \( y \):
   \[
   y = \frac{110}{(\sqrt[5]{12100})^2}
   \]

7. **Calculating z:**
   Once \( x \) and \( y \) are found, \( z \) can be calculated using the volume constraint \( z = \frac{110}{xy} \).

8. **Calculate the Cost:**
   Substitute the values of \( x \), \( y \), and \( z \) back into the cost function to find the minimum cost.

Due to the complexity of exact fifth roots, numerical methods or a calculator would be appropriate for finding \( x \) and \( y \) precisely.

Would you like more detailed steps or clarifications on any part? Here are some additional questions you might consider:

1. How do we solve the fifth root numerically?
2. Can we verify the critical points to confirm the minimum cost?
3. Are there any other constraints to consider in this optimization problem?
4. What are the practical applications of this optimization problem?
5. How does changing the cost per cm\(^2\) affect the overall minimum cost?

**Tip:** Always check for the reasonableness of your solution in optimization problems by verifying the critical points and ensuring they meet all the problem's constraints.

Discover how to minimize the cost of a rectangular box with a volume of 110 cm³, where top and bottom surfaces cost 8 cents per cm² and sides cost 8 cents per cm². Follow a detailed calculus-based solution for optimization.

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