We are tasked with finding the dimensions of a rectangular box with the largest volume given that the total surface area is 216 cm².

Step 1: Define variables

Let the dimensions of the rectangular box be:

• Length = $l$
• Width = $w$
• Height = $h$

The volume of the rectangular box is: $V = l \cdot w \cdot h$ The surface area of the rectangular box is: $S = 2lw + 2lh + 2wh = 216$ We want to maximize the volume $V = l \cdot w \cdot h$ subject to the constraint $2lw + 2lh + 2wh = 216$.

Step 2: Express one variable in terms of others

To simplify, let's assume the box is a cube, i.e., $l = w = h$, which is typical for maximizing volume in problems with fixed surface area. Then, the surface area equation becomes: $2l^2 + 2l^2 + 2l^2 = 216 \quad \Rightarrow \quad 6l^2 = 216$ Solving for $l$: $l^2 = \frac{216}{6} = 36 \quad \Rightarrow \quad l = \sqrt{36} = 6$ Since $l = w = h$, we conclude that: $\text{Length} = 6, \quad \text{Width} = 6, \quad \text{Height} = 6$

Final Answer:

• Length = 6
• Width = 6
• Height = 6

Would you like more details or have any questions?

Here are 5 related questions to expand the topic:

1. What is the volume of the box with the given dimensions?
2. How would the solution change if the surface area were 432 cm²?
3. What would happen to the volume if the box were not a cube?
4. How can you find the maximum volume if only two sides are equal?
5. How does surface area affect the volume of geometric shapes?

Tip: In problems involving maximum volume with constraints, checking symmetric shapes like cubes often simplifies the problem!