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To solve this problem, we need to minimize the surface area of an open-top rectangular box while keeping its volume constant at 4500 cubic centimeters (cc).

### Step 1: Define the dimensions
Let:
- \( w \) be the width of the box.
- \( l \) be the length of the box, and we are told \( l = 2w \) (the length is twice the width).
- \( h \) be the height of the box.

### Step 2: Volume equation
The volume \( V \) of a rectangular box is given by:
\[
V = l \times w \times h
\]
Substitute \( l = 2w \):
\[
V = 2w \times w \times h = 2w^2h
\]
We are told the volume is 4500 cc, so:
\[
2w^2h = 4500
\]
From this, solve for \( h \):
\[
h = \frac{4500}{2w^2} = \frac{2250}{w^2}
\]

### Step 3: Surface area equation
The surface area \( S \) of an open-top box (no top) is given by:
\[
S = l \times w + 2(l \times h) + 2(w \times h)
\]
Substitute \( l = 2w \):
\[
S = 2w \times w + 2(2w \times h) + 2(w \times h)
\]
Simplify:
\[
S = 2w^2 + 4wh + 2wh = 2w^2 + 6wh
\]
Now substitute \( h = \frac{2250}{w^2} \) from Step 2:
\[
S = 2w^2 + 6w \times \frac{2250}{w^2} = 2w^2 + \frac{13500}{w}
\]

### Step 4: Minimize the surface area
To minimize the surface area, take the derivative of \( S \) with respect to \( w \) and set it equal to zero:
\[
\frac{dS}{dw} = 4w - \frac{13500}{w^2}
\]
Set \( \frac{dS}{dw} = 0 \) and solve for \( w \):
\[
4w = \frac{13500}{w^2}
\]
Multiply both sides by \( w^2 \):
\[
4w^3 = 13500
\]
Solve for \( w \):
\[
w^3 = \frac{13500}{4} = 3375
\]
\[
w = \sqrt[3]{3375} = 15 \, \text{cm}
\]

### Step 5: Solve for height
Now that we have \( w = 15 \, \text{cm} \), substitute this into the expression for \( h \):
\[
h = \frac{2250}{w^2} = \frac{2250}{15^2} = \frac{2250}{225} = 10 \, \text{cm}
\]

### Conclusion:
The height that will make the surface area as small as possible is \( h = 10 \, \text{cm} \).

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Would you like further details on any step or more explanation?

Here are 5 related questions to explore further:
1. How would the problem change if the box had a top?
2. Can you calculate the minimum surface area for the box?
3. What happens if the length is three times the width instead of twice?
4. How does the surface area function change if the box’s volume is different?
5. How would this problem change if you wanted to minimize cost instead of surface area?

**Tip:** For optimization problems, always double-check that your critical points minimize (or maximize) the function by using the second derivative or other methods.

The volume of an open-top rectangular box is 4500 cc (cubic centimeters). The length of the rectangular base of the box is twice the width. What height will make the surface area as small as possible?

Learn how to minimize the surface area of an open-top rectangular box with a volume of 4500 cubic centimeters. Explore key concepts in optimization, algebra, and geometry as we calculate the optimal height. This solution is ideal for students in Grades 10-12.

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