Let's break down the problem to derive the volume $V$ of the container.

1. Dora starts with a rectangular piece of cardboard with dimensions 55 cm by 28 cm.
2. She cuts out squares with side length $x$ cm from each corner. After removing these squares, she folds the remaining flaps to form a box.

Step 1: Dimensions of the box after cutting

• The original dimensions of the cardboard are:

  • Length = 55 cm
  • Width = 28 cm

• After cutting out squares of side length $x$ cm from each corner, the new dimensions of the base of the box will be:

  • New length = $55 - 2x$ (since $x$ is removed from both ends)
  • New width = $28 - 2x$ (for the same reason)

Step 2: Volume of the box

The volume $V$ of the box is given by: $V = \text{length} \times \text{width} \times \text{height}$

• Length = $55 - 2x$
• Width = $28 - 2x$
• Height = $x$ (the side length of the squares cut out, as this becomes the height of the box)

Thus, the volume expression becomes: $V = (55 - 2x)(28 - 2x)(x)$

Step 3: Expanding the expression

Now, expand $(55 - 2x)(28 - 2x)$ using distributive property: $(55 - 2x)(28 - 2x) = 55 \times 28 - 55 \times 2x - 2x \times 28 + 2x \times 2x$ $= 1540 - 110x - 56x + 4x^2$ $= 1540 - 166x + 4x^2$

Now multiply this expression by $x$ (the height of the box): $V = (1540 - 166x + 4x^2) \times x$ $V = 1540x - 166x^2 + 4x^3$

Step 4: Final volume equation

Rearranging the terms gives the final volume expression: $V = 4x^3 - 166x^2 + 1540x$

This matches the given volume formula, $V = 4x^3 - 166x^2 + 1540x$.

Let me know if you'd like further clarification or details! Here are a few questions that might help extend your understanding:

1. What happens to the volume if $x$ becomes too large?
2. How do we find the optimal value of $x$ to maximize the volume of the box?
3. What are the dimensions of the base when $x = 5$ cm?
4. How does the volume change if the initial piece of cardboard is larger or smaller?
5. How can we use calculus to find the maximum volume of this box?

Tip: To optimize this box's volume, we can take the derivative of the volume function and find the critical points.