A box is constructed with a square top and bottom of side length x
. The material for the bottom of the box costs $7/cm2
and the material for the other five sides costs $4/cm2
. The box needs to be 11 cm3
.
Write an equation for the cost of constructing the box, in terms of x
:
C=
equation editorEquation Editor
Taking the derivative gives C′=
equation editorEquation Editor
The dimensions that minimize cost are side length
equation editorEquation Editor  cm
and height
equation editorEquation Editor  cm
.

To solve this problem, we need to first express the cost $ C $ of constructing the box in terms of the side length $ x $. Let's break this down step-by-step.

1. **Volume Constraint**:
   Given the volume of the box is 11 cm³, we have:
   \[
   V = x^2 h = 11 \implies h = \frac{11}{x^2}
   \]
   where $ h $ is the height of the box.

2. **Cost Calculation**:
   - The cost of the bottom: The bottom of the box has an area of $ x^2 $ and costs $7/cm². So, the cost of the bottom is:
     \[
     \text{Cost of bottom} = 7x^2
     \]
   - The cost of the top and four sides: The top and four sides together have a total area of:
     \[
     \text{Area of top} = x^2
     \]
     \[
     \text{Area of four sides} = 4 \times (x \cdot h) = 4xh = 4x \cdot \frac{11}{x^2} = \frac{44}{x}
     \]
     So, the total cost of the top and sides is:
     \[
     \text{Cost of top and sides} = 4 \cdot \left(x^2 + \frac{44}{x}\right) = 4x^2 + \frac{176}{x}
     \]
   - Total cost:
     \[
     C = \text{Cost of bottom} + \text{Cost of top and sides} = 7x^2 + 4x^2 + \frac{176}{x} = 11x^2 + \frac{176}{x}
     \]

So, the equation for the cost $ C $ is:
\[
C = 11x^2 + \frac{176}{x}
\]

3. **Finding the Derivative**:
   To find the dimensions that minimize the cost, we take the derivative of $ C $ with respect to $ x $ and set it to zero:
   \[
   C' = \frac{d}{dx} \left( 11x^2 + \frac{176}{x} \right) = 22x - \frac{176}{x^2}
   \]
   Setting $ C' = 0 $:
   \[
   22x - \frac{176}{x^2} = 0
   \]
   \[
   22x = \frac{176}{x^2}
   \]
   \[
   22x^3 = 176
   \]
   \[
   x^3 = 8
   \]
   \[
   x = 2 \text{ cm}
   \]

4. **Finding the Height**:
   Using $ x = 2 $ cm in the volume constraint equation $ h = \frac{11}{x^2} $:
   \[
   h = \frac{11}{2^2} = \frac{11}{4} = 2.75 \text{ cm}
   \]

So, the dimensions that minimize the cost are:
- Side length $ x = 2 $ cm
- Height $ h = 2.75 $ cm

### Summary
- Cost equation: $ C = 11x^2 + \frac{176}{x} $
- Derivative of cost: $ C' = 22x - \frac{176}{x^2} $
- Dimensions that minimize cost:
  - Side length $ x = 2 $ cm
  - Height $ h = 2.75 $ cm

Would you like more details or have any questions? 

Here are some related questions:
1. How do we verify if the critical point $ x = 2 $ cm is a minimum?
2. What is the second derivative of the cost function $ C $?
3. How does the cost change if the height constraint is modified?
4. What is the total cost of constructing the box with the dimensions found?
5. How does the cost function behave as $ x $ approaches 0 or $\infty$?
6. Can we generalize this cost minimization problem to other volumes?
7. What other shapes could potentially minimize material costs for a given volume?
8. How does changing the cost per unit area of materials affect the optimal dimensions?

**Tip**: When working on optimization problems, always check the second derivative to confirm whether the critical points are minima or maxima.

Learn how to optimize the construction cost of a box with a volume constraint using mathematical formulas and optimization techniques. Explore step-by-step solutions and key concepts like box volume, surface area cost calculation, and derivative applications. Suitable for advanced high school students and mathematics enthusiasts.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation