A box is constructed so that its length is twice its width. The sum of its length, width and depth is 140 cm. Calculate the values for its dimensions which result in a maximum surface area. Calculate this maximum surface area.
Step 1: Draw a diagram:
Step 2: Identify what needs to be optimised, and write an equation for this:
Step 3: Write an equation which relates x with y:
Step 4: Substitute for y:
Step 5: Differentiate and calculate maximum and/or minimum:
Step 6: Substitute for x into the equation from Step 4 to find the maximum/minimum:
Step 7: Calculate y and test your solution to make sure it works - some solutions don't: 
Step 8: Answer the question

### Step 1: Draw a Diagram
Imagine a rectangular box with dimensions labeled as:
- Length \( L \)
- Width \( W \)
- Depth \( D \)

Given that the length is twice the width, we have:
\[ L = 2W \]

### Step 2: Identify What Needs to Be Optimized, and Write an Equation for This
We need to maximize the surface area \( S \) of the box. The surface area \( S \) of a rectangular box is given by:
\[ S = 2(LW + LD + WD) \]

### Step 3: Write an Equation That Relates \( x \) with \( y \)
We are given that the sum of the length, width, and depth is 140 cm:
\[ L + W + D = 140 \]

Using the relationship \( L = 2W \), we substitute for \( L \):
\[ 2W + W + D = 140 \]
\[ 3W + D = 140 \]
\[ D = 140 - 3W \]

### Step 4: Substitute for \( y \)
Now, substitute \( D = 140 - 3W \) and \( L = 2W \) into the surface area equation:
\[
S = 2(LW + LD + WD) \\
S = 2(2W \cdot W + 2W \cdot (140 - 3W) + W \cdot (140 - 3W))
\]
Simplifying:
\[
S = 2(2W^2 + 280W - 6W^2 + 140W - 3W^2) \\
S = 2(-7W^2 + 420W)
\]
So, the equation for the surface area becomes:
\[
S = -14W^2 + 840W
\]

### Step 5: Differentiate and Calculate Maximum and/or Minimum
To find the maximum or minimum, differentiate \( S \) with respect to \( W \) and set the derivative equal to zero:
\[
\frac{dS}{dW} = -28W + 840
\]
Set \( \frac{dS}{dW} = 0 \):
\[
-28W + 840 = 0
\]
Solving for \( W \):
\[
W = \frac{840}{28} = 30 \text{ cm}
\]

### Step 6: Substitute \( W \) into the Equation from Step 4 to Find the Maximum
Substitute \( W = 30 \) cm back into the equation for \( D \) to find \( D \):
\[
D = 140 - 3(30) = 140 - 90 = 50 \text{ cm}
\]
Also, \( L = 2W = 2(30) = 60 \text{ cm} \).

### Step 7: Calculate \( y \) and Test Your Solution to Make Sure It Works
Now, let's calculate the maximum surface area using these dimensions:
\[
S = 2(60 \cdot 30 + 60 \cdot 50 + 30 \cdot 50)
\]
\[
S = 2(1800 + 3000 + 1500) = 2(6300) = 12600 \text{ cm}^2
\]
Thus, the maximum surface area is \( 12,600 \, \text{cm}^2 \).

### Step 8: Answer the Question
The dimensions of the box that maximize the surface area are:
- Length \( L = 60 \) cm
- Width \( W = 30 \) cm
- Depth \( D = 50 \) cm

The maximum surface area is \( 12,600 \, \text{cm}^2 \).

Would you like more details or have any questions?  
Here are some related questions:

1. How would the dimensions change if the sum of the dimensions was different?
2. What if the length was three times the width instead of two times?
3. Can you calculate the volume of the box with the optimized dimensions?
4. How would the surface area change if the depth were fixed at a specific value?
5. What other methods could be used to optimize the surface area?

**Tip:** Always check the second derivative to confirm whether you have a maximum or minimum when optimizing functions.

Explore how to maximize the surface area of a rectangular box with dimensions based on given constraints. Learn step-by-step how to optimize the surface area using mathematical equations and optimization techniques. Suitable for high school students.

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