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To solve this optimization problem, let's go through each part systematically. We are asked to minimize the surface area \( A(x) \) of a box with an open top, a square base with side length \( x \), and a given volume of 70304 cm³.

### Step 1: Find a formula for the surface area \( A(x) \) in terms of \( x \)

1. **Volume constraint**: Since the volume \( V \) of the box is given by:
   \[
   V = x^2 \cdot h = 70304
   \]
   where \( h \) is the height of the box. Solving for \( h \) in terms of \( x \),
   \[
   h = \frac{70304}{x^2}
   \]

2. **Surface area formula**: The surface area \( A(x) \) of the box, with an open top, consists of the base area \( x^2 \) and the four sides, each of area \( x \cdot h \):
   \[
   A(x) = x^2 + 4(x \cdot h)
   \]
   Substitute \( h = \frac{70304}{x^2} \):
   \[
   A(x) = x^2 + 4 \left( x \cdot \frac{70304}{x^2} \right)
   \]
   Simplifying,
   \[
   A(x) = x^2 + \frac{281216}{x}
   \]

### Step 2: Find the derivative \( A'(x) \)

To find \( A'(x) \), differentiate \( A(x) = x^2 + \frac{281216}{x} \) with respect to \( x \):
\[
A'(x) = 2x - \frac{281216}{x^2}
\]

### Step 3: Solve \( A'(x) = 0 \)

Set the derivative equal to zero to find the critical points:
\[
2x - \frac{281216}{x^2} = 0
\]
Multiply both sides by \( x^2 \) to clear the denominator:
\[
2x^3 = 281216
\]
Now, divide by 2:
\[
x^3 = 140608
\]
Take the cube root of both sides:
\[
x = \sqrt[3]{140608} \approx 52.48
\]

### Step 4: Verify that this value of \( x \) gives a minimum using the second derivative test

Find \( A''(x) \):
\[
A''(x) = 2 + \frac{2 \cdot 281216}{x^3} = 2 + \frac{562432}{x^3}
\]
Evaluate \( A''(x) \) at \( x \approx 52.48 \):
\[
A''(52.48) = 2 + \frac{562432}{(52.48)^3} > 0
\]
Since \( A''(x) > 0 \), this confirms that \( x \approx 52.48 \) gives a minimum for the surface area.

### Summary

- **Optimal side length \( x \):** Approximately 52.48 cm.
- **Height \( h \):** Using \( h = \frac{70304}{x^2} \), substitute \( x \approx 52.48 \):
  \[
  h \approx \frac{70304}{(52.48)^2} \approx 25.5 \text{ cm}
  \]
  
This minimizes the surface area of the box given the volume constraint.

Would you like more details on any of the steps, or do you have any questions?

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### Related Questions

1. What would happen to the optimal dimensions if the volume changed?
2. How would the solution differ if the box had a closed top?
3. What if the base was not square but rectangular?
4. How can we use Lagrange multipliers to solve this problem?
5. How does changing the number of open sides impact the solution?

### Tip

Always start by expressing one variable in terms of another using given constraints to simplify optimization problems.

A box with a square base and open top must have a volume of 70304 cm³. We wish to find the dimensions of the box that minimize the amount of material used. First, find a formula for the surface area of the box in terms of only x, the length of one side of the square base. Then find the derivative, set it to zero, and use the second derivative test to ensure it is a minimum.

Learn how to solve an optimization problem involving a box with a square base and open top that has a volume of 70304 cm³. This problem includes finding the dimensions that minimize the material used, calculating the derivative, setting it to zero, and applying the second derivative test to ensure a minimum. Suitable for college-level calculus students.

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