My brother wants to make an open-topped box out of a 4x 6 square foot piece of card board by cutting identical squares from the corners and folding up the sides. What is the dimension of each square he will cut out of each corner in order to max the volume of the box he makes.

To find the dimensions of each square to be cut out, we need to consider the volume of the box. The volume of a rectangular box is given by multiplying its length, width, and height.

Let's denote the length of the box as L, and the width as W. If square pieces of size x are cut out from each corner, then the length and width of the resulting box will be L-2x and W-2x respectively.

Now, we can calculate the volume of the box. Given that the piece of cardboard is 4x6 square feet, we can set up the equation:

Volume = (L-2x)(W-2x)x

Substituting L=4 and W=6 into the equation, we get:

Volume = (4-2x)(6-2x)x

To maximize the volume, we can differentiate this equation with respect to x and set the derivative equal to zero:

dV/dx = 8x^2 - 20x + 24 = 0

Now we can solve this quadratic equation to find the value of x that maximizes the volume. We can use the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a,

where a = 8, b = -20, and c = 24.

Plugging in these values, we get:

x = (-(-20) ± sqrt((-20)^2 - 4(8)(24))) / (2(8))

Simplifying further:

x = (20 ± sqrt(400 - 768)) / 16
x = (20 ± sqrt(-368)) / 16

Since the square root of a negative number is not real, we can conclude that the volume is maximized when x is a non-negative value. Therefore, x = 0.5 feet (or 6 inches) is the dimension of each square to be cut out.

To verify this, we can calculate the volume of the resulting box:

Volume = (4 - 2(0.5))(6 - 2(0.5))(0.5)
Volume = (3)(5)(0.5)
Volume = 7.5 cubic feet

Hence, cutting squares with dimensions of 0.5 feet (or 6 inches) from each corner will maximize the volume of the box.