An open top box is made by cutting congruent squares from the corners of a 12 inch by 9 inch sheet of cardboard and then folding the sides up to create the box. What are the dimensions of the box which contains the largest volume?
To find the dimensions of the box that will give us the largest volume, we need to understand the relationship between the dimensions of the cut squares and the resulting box.
Let's assume that the length of the cut square is x inches on each side. We will cut congruent squares from all four corners of the cardboard.
When we fold up the sides, the initial length of the box would be (12 - 2x) inches, and the initial width would be (9 - 2x) inches. The height of the box would be just the length of the cut squares, which is x inches.
To find the volume of the box, we simply multiply the length, width, and height together:
Volume = Length * Width * Height
Volume = (12 - 2x)(9 - 2x)(x)
Now, we have an equation for the volume of the box in terms of x. To find the value of x that will maximize the volume, we can take the derivative of the volume equation with respect to x and set it equal to zero:
dV/dx = 0
To solve this equation, let's simplify the equation:
dV/dx = (12 - 2x)(9 - 2x)(x)
Next, we will take the derivative of the simplified equation:
dV/dx = (12 - 2x)(9 - 2x) + (12 - 2x)(x) + (9 - 2x)(x)
Now, set the derivative equal to zero and solve for x:
(12 - 2x)(9 - 2x) + (12 - 2x)(x) + (9 - 2x)(x) = 0
Simplifying this equation further, we get:
(12 - 2x)(9 - 2x + x + x) = 0
(12 - 2x)(9 - x) = 0
Setting each factor in the equation equal to zero:
12 - 2x = 0 or 9 - x = 0
Solving for x:
12 - 2x = 0
2x = 12
x = 6
9 - x = 0
-x = -9
x = 9
We have two possible values for x, 6 inches and 9 inches. However, since the length of the cut squares cannot exceed half the length or width of the cardboard, which is 6 inches in this case, we discard the solution x = 9.
Therefore, the dimensions of the box that will give us the largest volume are:
Length = 12 - 2x = 12 - 2(6) = 12 - 12 = 0 inches
Width = 9 - 2x = 9 - 2(6) = 9 - 12 = -3 inches
Height = x = 6 inches
However, since a negative width does not make sense in this context, we need to reconsider our solution.
It seems that the dimensions we derived are not valid because the resulting box would have no length or negative width. This could mean that the maximum volume is not possible with the given dimensions.
To make sure, let's analyze the equation of volume again:
Volume = (12 - 2x)(9 - 2x)(x)
For the dimensions of the box to be valid, all three factors in the equation must be positive. The dimensions should be less than half the length and width of the cardboard (since we are cutting squares from each corner).
Let's calculate the maximum volume by substituting some values for x:
When x = 0 (no cut squares), the volume is 12 * 9 * 0 = 0.
When x = 1, the volume is 10 * 7 * 1 = 70.
When x = 2, the volume is 8 * 5 * 2 = 80.
When x = 3, the volume is 6 * 3 * 3 = 54.
From these calculations, we can see that the maximum volume occurs when x = 2, resulting in a volume of 80 cubic inches.
Therefore, the dimensions of the box that would give us the largest volume are:
Length = 12 - 2x = 12 - 2(2) = 12 - 4 = 8 inches
Width = 9 - 2x = 9 - 2(2) = 9 - 4 = 5 inches
Height = x = 2 inches
So, the dimensions of the box that would yield the largest volume are 8 inches by 5 inches by 2 inches.