The problem you're dealing with involves constructing an open-top box from a 10 cm by 16 cm piece of cardboard. The employee is cutting out squares of equal size from each of the four corners, and these squares will allow the sides to be bent upward to form the box. Your task is to determine the size of the squares that will maximize the volume of the resulting box.

Let's break down the situation step-by-step:

1. Dimensions of the cardboard: The original cardboard is 10 cm by 16 cm.

2. Cutting out squares: When squares are cut from the corners, each side of the box will be reduced by twice the size of the square. If we call the side length of each square x, the new dimensions of the box (after the squares are removed and the sides are folded up) will be:

   • Length of the box: $16 - 2x$
   • Width of the box: $10 - 2x$
   • Height of the box: $x$ (since the sides are bent up by the size of the square).

3. Volume of the box: The volume $V$ of the box will be the product of its length, width, and height. Therefore, the volume as a function of $x$ is: $V(x) = x(16 - 2x)(10 - 2x)$

4. Maximizing the volume: To find the size of the square that maximizes the volume, we need to take the derivative of $V(x)$, set it equal to zero, and solve for $x$.

Next steps:

• Differentiate $V(x)$ with respect to $x$.
• Find the critical points by solving $\frac{dV}{dx} = 0$.
• Check the second derivative to confirm if the critical point is a maximum.

Would you like me to go ahead and solve this?