We are tasked with determining the size of the squares to cut from the corners of a 10-inch by 16-inch piece of cardboard, which will allow us to create a rectangular box with an open top and the largest possible volume.

Let's set up the problem:

1. The cardboard is 10 inches by 16 inches.
2. Small squares of side length x inches are cut from each corner.
3. After cutting and folding up the sides, the dimensions of the base of the box will be:
   • Length: $16 - 2x$ inches
   • Width: $10 - 2x$ inches
4. The height of the box will be $x$ inches (the size of the square cut and folded).

Thus, the volume V of the box is given by: $V(x) = x(16 - 2x)(10 - 2x)$ We need to maximize this volume.

Steps to solve:

1. Expand the volume function.
2. Find the critical points by taking the derivative of the volume function and setting it equal to zero.
3. Determine the value of x that maximizes the volume by solving the equation and confirming it with the second derivative test.

Let me compute the optimal x for maximum volume.The critical points for the volume function are x = 2 and x = 20/3 (which is approximately 6.67). However, since the box dimensions must be positive, $x = 20/3$ would result in negative base dimensions, making it an invalid solution.

Thus, the valid solution is $x = 2$.

This means the squares cut from each corner should have a side length of 2 inches to maximize the volume of the box.

Would you like further details on this calculation, or perhaps help with related questions?

Related questions:

1. What is the maximum volume achieved with $x = 2$?
2. What happens to the volume if a different size of square is cut?
3. Can you solve this using a different approach, like trial and error?
4. How does the second derivative confirm that this is a maximum?
5. What would the dimensions of the box be after cutting?

Tip:

In optimization problems, always check boundary conditions and critical points to ensure you're maximizing or minimizing correctly.