If 1500 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box

To find the largest possible volume of a box with a square base and an open top using 1500 square centimeters of material, we need to set up and solve an optimization problem.

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### Step 1: Define variables
- Let \( x \) be the length of one side of the square base.
- Let \( h \) be the height of the box.

---

### Step 2: Constraints and relationships
1. **Surface area constraint**:
   The box has:
   - A square base with an area \( x^2 \),
   - Four rectangular sides, each with an area \( xh \).

   Total surface area is:
   \[
   x^2 + 4xh = 1500
   \]

2. **Volume of the box**:
   The volume \( V \) is:
   \[
   V = x^2h
   \]

---

### Step 3: Solve for \( h \) in terms of \( x \)
From the surface area equation:
\[
h = \frac{1500 - x^2}{4x}
\]

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### Step 4: Substitute \( h \) into the volume equation
\[
V = x^2 \cdot \frac{1500 - x^2}{4x}
\]
Simplify:
\[
V = \frac{x(1500 - x^2)}{4}
\]
\[
V = \frac{1500x - x^3}{4}
\]

---

### Step 5: Maximize \( V \)
Differentiate \( V \) with respect to \( x \):
\[
\frac{dV}{dx} = \frac{1500 - 3x^2}{4}
\]

Set \( \frac{dV}{dx} = 0 \):
\[
1500 - 3x^2 = 0
\]
\[
x^2 = 500
\]
\[
x = \sqrt{500} = 10\sqrt{5} \approx 22.36 \, \text{cm}
\]

---

### Step 6: Determine \( h \)
Substitute \( x = 10\sqrt{5} \) into the equation for \( h \):
\[
h = \frac{1500 - (10\sqrt{5})^2}{4 \cdot 10\sqrt{5}}
\]
\[
h = \frac{1500 - 500}{40\sqrt{5}}
\]
\[
h = \frac{1000}{40\sqrt{5}} = \frac{25}{\sqrt{5}} = 5\sqrt{5} \, \text{cm}
\]

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### Step 7: Find the maximum volume
\[
V = x^2h = (10\sqrt{5})^2 \cdot 5\sqrt{5}
\]
\[
V = 500 \cdot 5\sqrt{5} = 2500\sqrt{5} \, \text{cm}^3
\]
\[
V \approx 5590.17 \, \text{cm}^3
\]

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### Final Answer:
The largest possible volume of the box is approximately **5590.17 cm³**.

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Would you like detailed explanations of any specific step?  

#### Related Questions:
1. What would happen to the volume if the material were doubled?
2. How does the height of the box change if the base size changes?
3. What is the relationship between surface area and volume for this shape?
4. How do derivatives help in finding maximum values in such problems?
5. Could the box have a closed top with the same material constraint?

#### Tip:
Always check the constraints carefully when solving optimization problems, as they dictate feasible solutions.

Discover how to find the largest possible volume of a box with a square base and open top, given 1500 cm² of material. This step-by-step guide uses optimization techniques, derivatives, and constraints, ideal for Grades 11-12 calculus students.

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