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? use V = x (24-2x)(24-2x)
To find 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, we can use the formula V = x(24-2x)(24-2x), where x represents the side length of the squares being cut.
The first step is to find the value of x that maximizes the volume. We can do this by finding the critical points of the function V = x(24-2x)(24-2x) and determining which one corresponds to a maximum.
To find the critical points, we can take the derivative of V with respect to x:
dV/dx = (24-2x)(24-2x) + x(-2)(24-2x) + x(24-2x)(-2)
= (24-2x)(24-2x) - 4x(24-2x) - 4x(24-2x)
= (24-2x)(24-2x) - 8x(24-2x)
Next, we set the derivative equal to 0 and solve for x:
(24-2x)(24-2x) - 8x(24-2x) = 0
Simplifying this equation, we get:
576 - 96x + 4x^2 - 192x + 32x^2 - 64x^2 = 0
- 24x^2 + 4x^2 - 96x - 192x - 64x^2 + 576 = 0
- 84x^2 - 288x + 576 = 0
Now, we can solve this quadratic equation for x. We can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
For this equation, a = -84, b = -288, and c = 576. Plugging these values into the quadratic formula, we get:
x = (-(-288) ± sqrt((-288)^2 - 4(-84)(576))) / (2(-84))
x = (288 ± sqrt(82944 + 192768)) / 168
x = (288 ± sqrt(275712)) / 168
x = (288 ± 524.63) / 168
x ≈ 4.26 or x ≈ 4.81
Since the length of the side must be positive, we can ignore the negative solution. Therefore, the value of x that maximizes the volume is approximately 4.81 inches.
To find the volume of the largest box, we substitute this value of x into the formula V = x(24-2x)(24-2x):
V ≈ 4.81(24-2(4.81))(24-2(4.81))
V ≈ 4.81(24-9.62)(24-9.62)
V ≈ 4.81(14.38)(14.38)
V ≈ 978.075
Therefore, the volume of the largest box that can be made from the square piece of cardboard is approximately 978.075 cubic inches.