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The problem provided is about finding the dimensions of an open-top box with a square base that minimizes the surface area, subject to a fixed volume of \(4 \, \text{m}^3\). The given solution suggests that the dimensions are \(x = 2 \, \text{m}\) (the side of the square base), \(y = 1 \, \text{m}\) (the height of the box), and the minimum surface area is \(12 \, \text{m}^2\).

Let’s walk through how to solve this step by step:

### Variables:
- \(x\): Side length of the square base (in meters).
- \(y\): Height of the box (in meters).
- The volume of the box: \(V = 4 \, \text{m}^3\).

### Step 1: Volume Constraint
The volume of the box is the area of the base times the height:
\[
V = x^2 y = 4 \, \text{m}^3
\]
So, we have:
\[
x^2 y = 4
\]
or equivalently,
\[
y = \frac{4}{x^2}
\]
This gives us the height in terms of the base side length \(x\).

### Step 2: Surface Area to Minimize
The surface area \(A\) of the box consists of the area of the square base and the four vertical sides (since the box has no top):
\[
A = x^2 + 4xy
\]
Substitute \(y = \frac{4}{x^2}\) from the volume constraint into this equation:
\[
A(x) = x^2 + 4x \cdot \frac{4}{x^2} = x^2 + \frac{16}{x}
\]
So, the goal is to minimize this function \(A(x) = x^2 + \frac{16}{x}\).

### Step 3: Minimize the Surface Area
To minimize \(A(x)\), take the derivative of \(A(x)\) with respect to \(x\):
\[
\frac{dA}{dx} = 2x - \frac{16}{x^2}
\]
Set the derivative equal to zero to find the critical points:
\[
2x - \frac{16}{x^2} = 0
\]
Multiply through by \(x^2\) to eliminate the denominator:
\[
2x^3 = 16
\]
Solve for \(x\):
\[
x^3 = 8 \quad \Rightarrow \quad x = 2
\]

### Step 4: Find the Height \(y\)
Now that we know \(x = 2 \, \text{m}\), use the volume equation to find \(y\):
\[
y = \frac{4}{x^2} = \frac{4}{2^2} = 1 \, \text{m}
\]

### Step 5: Minimum Surface Area
Now calculate the minimum surface area:
\[
A = x^2 + \frac{16}{x} = 2^2 + \frac{16}{2} = 4 + 8 = 12 \, \text{m}^2
\]

### Final Answer:
- Dimensions: \(x = 2 \, \text{m}\) (side of the square base), \(y = 1 \, \text{m}\) (height).
- Minimum surface area: \(A = 12 \, \text{m}^2\).

Would you like more details on any step, or do you have any other questions?

### Related Questions:
1. How does changing the volume constraint affect the minimum surface area?
2. Can the same method be used if the base is rectangular, not square?
3. What happens to the surface area if the box is closed at the top?
4. Can the box have fractional dimensions for the base or height in this context?
5. How do you apply the second derivative test to confirm a minimum for this problem?

### Tip:
Always verify critical points by checking the second derivative or by analyzing the behavior of the function around those points to ensure a minimum, not a maximum.

Find the dimensions that use the minimum amount of material to construct an open box with square base and that has a volume of 4 m³.

Learn how to find the dimensions of an open box with a square base that minimizes material usage while having a fixed volume of 4 m³. This calculus optimization problem involves using derivatives to minimize the surface area.

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