We are tasked with finding the dimensions of a box that result in the maximum surface area. Here's how to approach it step by step.

Step 1: Define Variables

Let:

• $l$ be the length of the box (in cm),
• $w$ be the width of the box (in cm),
• $d$ be the depth (height) of the box (in cm).

From the problem:

1. The length is twice the width: $l = 2w$
2. The sum of the length, width, and depth is 140 cm: $l + w + d = 140$ Substituting $l = 2w$ into this equation gives: $2w + w + d = 140$ Simplifying: $3w + d = 140 \quad \Rightarrow \quad d = 140 - 3w$

Step 2: Write the Surface Area Function

The surface area $A$ of a box is given by the formula: $A = 2lw + 2lh + 2wh$ Substituting $l = 2w$ and $d = 140 - 3w$ into this equation gives: $A = 2(2w)(w) + 2(2w)(140 - 3w) + 2w(140 - 3w)$ Simplifying each term: $A = 4w^2 + 4w(140 - 3w) + 2w(140 - 3w)$ $A = 4w^2 + 560w - 12w^2 + 280w - 6w^2$ $A = -14w^2 + 840w$

Step 3: Maximize the Surface Area

We now have the surface area function: $A(w) = -14w^2 + 840w$ To maximize $A$, take the derivative of $A$ with respect to $w$, and set it equal to zero: $\frac{dA}{dw} = -28w + 840$ Set $\frac{dA}{dw} = 0$ to find the critical points: $-28w + 840 = 0 \quad \Rightarrow \quad w = \frac{840}{28} = 30$

Step 4: Find the Corresponding Dimensions

• Width $w = 30 \, \text{cm}$
• Length $l = 2w = 2(30) = 60 \, \text{cm}$
• Depth $d = 140 - 3w = 140 - 3(30) = 50 \, \text{cm}$

Step 5: Calculate the Maximum Surface Area

Now, calculate the maximum surface area using these dimensions: $A = 2lw + 2lh + 2wh = 2(60)(30) + 2(60)(50) + 2(30)(50)$ $A = 2(1800) + 2(3000) + 2(1500)$ $A = 3600 + 6000 + 3000 = 12,600 \, \text{cm}^2$

Conclusion:

• The dimensions of the box that result in the maximum surface area are:
  • Length $l = 60 \, \text{cm}$
  • Width $w = 30 \, \text{cm}$
  • Depth $d = 50 \, \text{cm}$
• The maximum surface area is $12,600 \, \text{cm}^2$.

Would you like more details or have any questions?

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Related Questions:

1. What is the volume of the box with these dimensions?
2. How does the surface area change if the depth is fixed instead of the sum constraint?
3. What happens to the surface area if the width is halved?
4. How would the calculations change if the length was thrice the width?
5. Can we achieve a maximum volume for this box under similar constraints?

Tip:

When solving optimization problems, always remember to verify whether you are maximizing or minimizing by checking the second derivative.