Congruent squares are cut from the corners of a 20 in. square piece of tine, and the edges are turned up to make an open box. How large should the squares cut from the corners be in order to maximize the volume of the box?

v = x(20-2x)^2

v' = 4(3x^2 - 40x + 100)
v'=0 when x=10/3

To maximize the volume of the box, we need to find the optimal size of the squares cut from the corners. Let's break down the problem into steps:

1. Visualize the problem: Before we start, it's helpful to have a clear understanding of the situation. Imagine a 20 in. square piece of tin. We will cut congruent squares from each of the corners and fold up the edges to create the box.

2. Define variables: Let's denote the side length of the square cut from the corners as x (in inches).

3. Determine the dimensions of the resulting box: When we fold up the edges of the remaining shape, we will create a box with a rectangular base and height.

- The rectangular base will have a length of 20 - 2x inches (the remaining side after cutting the squares from the corners) and a width of 20 - 2x inches.
- The height of the box will be the side length of the squares cut from the corners, which is x inches.

4. Express the volume: The volume of a rectangular box can be calculated by multiplying its length, width, and height. In this case, the volume is given by:
V = (20 - 2x)(20 - 2x)x
V = x(20 - 2x)(20 - 2x)
V = 4x(10 - x)(10 - x) (after simplification)

5. Maximize the volume: To find the maximum volume, we need to find the value of x that maximizes the expression V = 4x(10 - x)(10 - x). We can do this by taking the derivative of V with respect to x, setting it equal to zero, and solving for x.
dV/dx = 4(10 - 2x)(10 - x) + 4x(-1)(10 - x) (using the product rule)
Setting dV/dx equal to zero:
4(10 - 2x)(10 - x) + 4x(-1)(10 - x) = 0
Simplifying and solving for x will give us the optimal value.

6. Calculate the optimal value: By solving the equation derived in step 5, we can find the value of x that maximizes the volume. This value represents the side length of the small squares that should be cut from the corners.

Now, you can go ahead and solve the equation to find the value of x, which will give you the optimal size of the squares to be cut from the corners in order to maximize the volume of the box.