An open-top box is to be constructed from a 6 foot by 8 foot rectangular cardboard by cutting out equal squares at each corner and the folding up the flaps. Let x denote the length of each side of the square to be cut out.
a)
To find the length of each side of the square to be cut out, we can start by visualizing the problem and understanding what needs to be done.
1. Visualize the rectangular cardboard: We have a rectangular cardboard with dimensions 6 feet by 8 feet. Let's draw a rectangle on a piece of paper or imagine it in our mind.
2. Cut out squares: We need to cut equal squares from each corner of the rectangular cardboard. The side length of these squares is denoted by x. Draw squares with side lengths x on all four corners of the rectangle.
3. Fold up the flaps: After cutting out the squares, we fold up the flaps to form the open-top box. The flaps will have a height equal to the side length of the square, which is x.
Now, let's analyze the dimensions of the open-top box that we have created:
Length: The length of the box will be the original length of the cardboard minus the lengths of the two flaps on each side. Since we cut out squares on both ends of the length, the length of the box will be (8 - 2x) feet.
Width: Similarly, the width of the box will be the original width of the cardboard minus the lengths of the two flaps on each side. Since we cut out squares on both ends of the width, the width of the box will be (6 - 2x) feet.
Height: The height of the box will be the length of the flaps, which is x feet.
Now we have the dimensions of the open-top box in terms of x. The goal is to find the value of x that will maximize the volume of the box.
To find the maximum volume, we use the fact that the volume of a rectangular box is given by multiplying its length, width, and height.
Volume = Length * Width * Height
= (8 - 2x) * (6 - 2x) * x
To maximize the volume, we can take the derivative of the volume function with respect to x, set it equal to zero, and solve for x.
Let's calculate the derivative:
dV/dx = (8 - 2x)(6 - 2x) + (8 - 2x)(-2x) + (6 - 2x)(-2x)
Simplifying the equation and setting it equal to zero:
0 = (8 - 2x)(6 - 2x) + (8 - 2x)(-2x) + (6 - 2x)(-2x)
Now, we can solve this equation for x to find the value that maximizes the volume.