The problem asks to find the maximum volume of a box with no top, which is formed by cutting squares out of the corners of a rectangular cardboard and folding up the sides. Here's the breakdown of how to solve it:

Step 1: Define variables

Let $x$ represent the side length of the square cut from each corner.

• The dimensions of the original rectangular piece of cardboard are 25 inches by 48 inches.
• After cutting out squares from the corners, the new dimensions of the base of the box will be:
  • Length = $48 - 2x$
  • Width = $25 - 2x$
  • Height = $x$ (the height of the box comes from the square cut and folding the sides up).

Step 2: Volume formula

The volume $V$ of the box is given by the formula: $V(x) = \text{length} \times \text{width} \times \text{height} = (48 - 2x)(25 - 2x)(x)$ This simplifies to: $V(x) = x(48 - 2x)(25 - 2x)$

Step 3: Maximizing the volume

To maximize the volume, we need to take the derivative of $V(x)$ with respect to $x$, set it equal to 0, and solve for $x$. Let's first expand the volume expression:

$V(x) = x[(48)(25) - 2(48)x - 2(25)x + 4x^2]$ $V(x) = x[1200 - 96x - 50x + 4x^2]$ $V(x) = x(1200 - 146x + 4x^2)$ $V(x) = 1200x - 146x^2 + 4x^3$

Now, take the derivative $V'(x)$:

$V'(x) = 1200 - 292x + 12x^2$

Set $V'(x) = 0$ to find the critical points:

$1200 - 292x + 12x^2 = 0$

I will now solve this quadratic equation to find the values of $x$.The critical points are: