What is the volume of the largest box that can be made from a square piece of cardboard with side lengths of 24 inches by cutting equal squares from each corner and turning up the sides?
To find the volume of the largest box, we need to maximize the height of the box.
Let's start by visualizing the process of cutting equal squares from each corner and turning up the sides.
By cutting the same size squares from each corner, we create flaps on all four sides of the square piece of cardboard. When we fold up these flaps, we form the sides of the box, with the corners of the square piece of cardboard becoming the base of the box.
Since the flaps have an equal width and length, and the original side length of the square piece of cardboard is 24 inches, each flap has a width and length equal to the side length of the square cut out from each corner.
Let's say the side length of the square cut out from each corner is x inches.
The width of the box is then equal to the side length of the original square, minus twice the length of each flap, which is 24 - 2x inches.
Similarly, the length of the box is also equal to 24 - 2x inches.
The height of the box is equal to the length of each flap, which is x inches.
To find the volume of the box, we multiply the length, width, and height:
Volume = length * width * height
= (24 - 2x) * (24 - 2x) * x
= x(24 - 2x)(24 - 2x)
To maximize the volume, we need to find the value of x that results in the maximum value for the expression x(24 - 2x)(24 - 2x).
To do this, we can find the derivative of the expression with respect to x and set it equal to zero. The critical points we find will be either maximum or minimum points.
Let's find the derivative:
dV/dx = (24 - 2x)(24 - 2x) + x(-2)(24 - 2x) + x(24 - 2x)(-2)
= (24 - 2x)^2 - 4x(24 - 2x)
Setting the derivative equal to zero:
(24 - 2x)^2 - 4x(24 - 2x) = 0
Expanding and simplifying:
(24 - 2x)(24 - 2x) - 4x(24 - 2x) = 0
(576 - 48x + 4x^2) - (96x - 8x^2) = 0
576 - 48x + 4x^2 - 96x + 8x^2 = 0
12x^2 - 144x + 576 = 0
x^2 - 12x + 48 = 0
Using the quadratic formula:
x = (-(-12) ± sqrt((-12)^2 - 4(1)(48))) / (2(1))
x = (12 ± sqrt(144 - 192)) / 2
x = (12 ± sqrt(-48)) / 2
The expression inside the square root is negative, which means there are no real solutions for x.
However, we can still determine the maximum volume by considering the endpoints of the range we're interested in. Since we're cutting squares from each corner, the side length of those squares cannot exceed half the original side length of the square piece of cardboard. Therefore, the range for x is 0 ≤ x ≤ 12.
Considering the endpoints of the range, when x = 0 or x = 12, the volume of the box would be 0.
Therefore, the maximum volume of the box can only be achieved when x is a number between 0 and 12. Though we know the precise point of maximum volume lies within this interval based on the calculus steps, we can continue to find the maximum volume by evaluating the volume expression at the endpoints:
Volume when x = 0:
V = 0(24 - 2(0))(24 - 2(0))
V = 0
Volume when x = 12:
V = 12(24 - 2(12))(24 - 2(12))
V = 12(24 - 24)(24 - 24))
V = 0
As stated before, the maximum volume can only be achieved when x is a number between 0 and 12. Therefore, the largest volume of the box that can be made from a square piece of cardboard is 0 cubic inches.