Minimizing Packaging Costs If an open box has a square base and a volume of 103 in.3 and is constructed from a tin sheet, find the dimensions of the box, assuming a minimum amount of material is used in its construction. (Round your answers to two decimal places.)
height	
  in
length	
  in
width	
  in

To minimize the amount of material used in constructing an open box with a square base and a given volume, we need to minimize the surface area subject to the constraint of a fixed volume. Here's how we can approach this:

### Given:
- The volume of the box: \( V = 103 \, \text{in}^3 \)
- The base is square with side length \( x \), and the height of the box is \( h \).
  
### Step 1: Express the volume constraint
The volume \( V \) of the box is given by:
\[
V = x^2 h
\]
We know \( V = 103 \), so:
\[
x^2 h = 103
\]
This gives us the relationship between \( x \) and \( h \):
\[
h = \frac{103}{x^2}
\]

### Step 2: Express the surface area
Since the box has no top, the surface area \( A \) consists of the area of the base and the four sides. The surface area is:
\[
A = x^2 + 4xh
\]
Substitute \( h \) from the volume constraint into the surface area formula:
\[
A = x^2 + 4x \left( \frac{103}{x^2} \right)
\]
Simplify this expression:
\[
A = x^2 + \frac{412}{x}
\]

### Step 3: Minimize the surface area
To minimize \( A \), we take the derivative of \( A \) with respect to \( x \), set it equal to zero, and solve for \( x \).

First, find the derivative:
\[
\frac{dA}{dx} = 2x - \frac{412}{x^2}
\]
Set this equal to zero to find the critical points:
\[
2x - \frac{412}{x^2} = 0
\]
Multiply both sides by \( x^2 \) to eliminate the fraction:
\[
2x^3 = 412
\]
Solve for \( x \):
\[
x^3 = 206
\]
\[
x = \sqrt[3]{206} \approx 5.88 \, \text{in}
\]

### Step 4: Find the height \( h \)
Now that we have \( x \), we can find the height \( h \) using the volume equation:
\[
h = \frac{103}{x^2} = \frac{103}{(5.88)^2} \approx 2.98 \, \text{in}
\]

### Final Dimensions:
- Height: \( h \approx 2.98 \, \text{in} \)
- Length: \( x \approx 5.88 \, \text{in} \)
- Width: \( x \approx 5.88 \, \text{in} \)

Would you like more details on any specific step?

### Related questions:
1. How do we confirm that the critical point is a minimum?
2. How would the process change if the box had a lid?
3. How does the material cost change if the volume is increased?
4. What if the base of the box is rectangular, not square?
5. How would the dimensions change if the surface area were fixed instead of the volume?

### Tip:
When minimizing or maximizing quantities in constrained optimization problems, always express one variable in terms of the other to reduce the number of variables in your equations.

Minimizing Packaging Costs: If an open box has a square base and a volume of 103 in³ and is constructed from a tin sheet, find the dimensions of the box, assuming a minimum amount of material is used in its construction. (Round your answers to two decimal places.)

Learn how to minimize the packaging costs of an open box with a square base and a fixed volume of 103 in³ using calculus. This problem covers optimization techniques involving surface area and volume. The solution provides detailed steps for calculating the dimensions of the box that minimize the material used, making it suitable for high school students in Grades 11-12.

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