A square of size x inches is cut out of each corner of an 8in by 12in piece of cardboard, and the sides are folded up to form an open-topped box. Determine the dimensions of the cut-out squares that will produce the box of maximum volume.
To find the dimensions of the cut-out squares that will produce the box of maximum volume, we need to follow a few steps:
Step 1: Define the variables
Let's assume that the side length of the cut-out square is "y" inches.
Step 2: Calculate the dimensions of the base of the box
Since we cut out squares from each corner, the base of the box would have dimensions (8 - 2y) inches by (12 - 2y) inches.
Step 3: Calculate the height of the box
The height of the box would be equal to the side length of the cut-out square, which is "y" inches.
Step 4: Calculate the volume of the box
The volume of the box can be calculated using the formula: Volume = length * width * height. Substituting the values, we get:
Volume = (8 - 2y) * (12 - 2y) * y
Step 5: Simplify the volume equation
Expand the equation and simplify:
Volume = (96y - 40y^2 + 4y^3) cubic inches
Step 6: Find the derivative of the volume equation
To find the maximum volume, we need to find the critical points of the volume equation. Take the derivative of the volume equation with respect to "y":
dV/dy = 96 - 80y + 12y^2
Step 7: Set the derivative equal to zero and solve for "y"
Setting the derivative equal to zero gives us:
96 - 80y + 12y^2 = 0
Step 8: Solve the quadratic equation
Solve the quadratic equation using factoring, completing the square, or the quadratic formula. In this case, the equation can be factored as follows:
12y^2 - 80y + 96 = 0
3y^2 - 20y + 24 = 0
(3y - 8)(y - 3) = 0
So, the possible values for "y" are y = 8/3 or y = 3.
Step 9: Evaluate the critical points
Evaluate the critical points by substituting back into the volume equation:
For y = 8/3, Volume = (8 - 2(8/3))(12 - 2(8/3))(8/3) = 256/27 cubic inches
For y = 3, Volume = (8 - 2(3))(12 - 2(3))(3) = 108 cubic inches
Step 10: Determine the maximum volume
Comparing the two volumes, we find that the maximum volume occurs when y = 3. Therefore, the dimensions of the cut-out squares that will produce the box of maximum volume are 3 inches by 3 inches.