The volume of an open-top rectangular box is 3500 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? 

Let’s solve the problem step by step.

### Step 1: Define Variables
Let:
- \( w \) be the width of the base of the box (in cm).
- \( l \) be the length of the base of the box (in cm). Given that the length is twice the width, we have \( l = 2w \).
- \( h \) be the height of the box (in cm).

### Step 2: Volume Equation
The volume \( V \) of the box is given by:

\[
V = l \cdot w \cdot h = 2w \cdot w \cdot h = 2w^2h
\]

We know the volume is 3500 cc, so:

\[
2w^2h = 3500
\]

Solving for \( h \):

\[
h = \frac{3500}{2w^2} = \frac{1750}{w^2}
\]

### Step 3: Surface Area Equation
The surface area \( A \) of the open-top box (no lid) is given by:

\[
A = lw + 2lh + 2wh
\]

Substituting \( l = 2w \):

\[
A = (2w \cdot w) + 2(2w \cdot h) + 2(w \cdot h) = 2w^2 + 4wh + 2wh = 2w^2 + 6wh
\]

Substituting \( h = \frac{1750}{w^2} \):

\[
A = 2w^2 + 6w \left(\frac{1750}{w^2}\right) = 2w^2 + \frac{10500}{w}
\]

### Step 4: Minimize the Surface Area
To find the height that minimizes the surface area, we need to find the critical points of \( A \). We do this by taking the derivative of \( A \) with respect to \( w \) and setting it to zero:

\[
\frac{dA}{dw} = 4w - \frac{10500}{w^2}
\]

Set the derivative equal to zero to find the critical points:

\[
4w - \frac{10500}{w^2} = 0
\]

\[
4w^3 = 10500
\]

\[
w^3 = \frac{10500}{4} = 2625
\]

\[
w = \sqrt[3]{2625}
\]

Calculating \( w \):

\[
w \approx 13.64 \, \text{cm}
\]

### Step 5: Find the Height \( h \)
Now, use the value of \( w \) to find \( h \):

\[
h = \frac{1750}{w^2} = \frac{1750}{(13.64)^2}
\]

Calculating \( h \):

\[
h \approx 9.4 \, \text{cm}
\]

So, the height \( h \) that makes the surface area as small as possible is approximately \( 9.4 \) cm.

Would you like more details or have any questions?

### Related Questions:
1. How do you find the critical points of a function?
2. How does the width of the base affect the surface area of the box?
3. What would happen if the box had a lid? How would that change the calculations?
4. How do you use the second derivative test to confirm if a critical point is a minimum or maximum?
5. How does changing the volume of the box affect the optimal height?

**Tip:** For optimization problems, always ensure your function is correctly defined and includes all constraints before finding the critical points!

The volume of an open-top rectangular box is 3500 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?

Solve an optimization problem involving an open-top rectangular box with a volume of 3500 cc. The length is twice the width, and we calculate the height that minimizes the surface area. This problem covers key concepts like volume, surface area, and using derivatives to find critical points. Suitable for Grades 10-12.

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