A square piece of cardboard, 24 inches by 24 inches, is to be made into an open box by cutting out each of the four corners. Each side of the box will then be folded up. Find the maximum volume that the box can hold.
let the size of the cut-out be x by x inches
length of base = 24-2x
width of base = 24-2x
height = x
V = x(24-x)^2
= x(576 - 48x + x^2)
=576x - 48x^2 + x^2
dV/dx = 576 - 96x + 3x^2 = 0 for a max of V
x^2 - 32x + 192 = 0
(x - 24)(x - 8) = 0
x - 8 or x = 24, but if x = 24, we have a negative side
so x = 8
V = 8(24-16)^2 = 512 inches^3
To find the maximum volume that the box can hold, we need to determine the dimensions of the box after cutting and folding.
Let's denote x as the length of each side to be cut from each corner.
After cutting off the corners, the dimensions of the box will be (24-2x) inches by (24-2x) inches by x inches.
The volume of the box can be calculated by multiplying the three dimensions:
V(x) = (24-2x)(24-2x)(x)
To find the maximum volume, we need to find the value of x that maximizes V(x).
Taking the derivative of V(x) with respect to x:
V'(x) = 4(x-12)(x-12)
Setting V'(x) to 0 and solving for x:
4(x-12)(x-12) = 0
(x-12)(x-12) = 0
x-12 = 0
x = 12
The critical point is x = 12.
To determine whether it is a maximum or minimum, we can evaluate the second derivative of V(x):
V''(x) = 12x - 48
Plug in the critical point:
V''(12) = 12(12) - 48 = 144 - 48 = 96
Since V''(12) is positive, the critical point x = 12 is a minimum.
Therefore, the maximum volume of the box can be obtained by setting x = 12:
V(12) = (24-2(12))(24-2(12))(12)
V(12) = (0)(0)(12)
V(12) = 0
Hence, the maximum volume that the box can hold is 0 cubic inches.
To find the maximum volume that the box can hold, we need to determine the dimensions of the cut-out corners that will result in the largest possible volume.
Let's assume that each side of the cut-out corner has length "x" inches.
We know that the original length and width of the cardboard are both 24 inches, so after cutting out the corners, the length of the box will be 24 - 2x inches, and the width of the box will also be 24 - 2x inches.
The height of the box will be equal to the length of the cut-out corner, which is x inches.
Therefore, the volume of the box can be calculated as V = length * width * height.
Substituting the values, we have:
V = (24 - 2x)(24 - 2x)(x)
To find the maximum volume, we need to maximize this equation. We can do this by finding the critical points of the equation, which are the points where the derivative is equal to zero.
Let's differentiate the equation with respect to x:
dV/dx = 4(24 - 2x)(x) + (24 - 2x)(24 - 2x) = 4x(24 - 2x) + (24 - 2x)(24 - 2x)
Now, set the derivative equal to zero and solve for x:
0 = 4x(24 - 2x) + (24 - 2x)(24 - 2x)
Simplifying the equation:
0 = 4x(24 - 2x) + (576 - 48x - 48x + 4x^2)
0 = 4x(24 - 2x) + 4x^2 - 96x + 576
0 = 4x^2 - 96x + 576
Now, we can solve this quadratic equation to find the critical points:
4x^2 - 96x + 576 = 0
Dividing both sides by 4:
x^2 - 24x + 144 = 0
Now, solve the quadratic equation using factoring, completing the square, or using the quadratic formula. In this case, factoring is the most convenient:
(x - 12)(x - 12) = 0
So, x = 12
Now, we need to check if this critical point gives us a maximum volume. To do that, we can compute the second derivative:
d^2V/dx^2 = 12 - 48 = -36
Since the second derivative is negative, the critical point x = 12 indeed gives us a maximum volume.
Finally, substitute this value of x into the volume equation to find the maximum volume:
V = (24 - 2x)(24 - 2x)(x)
V = (24 - 2(12))(24 - 2(12))(12)
V = (24 - 24)(24 - 24)(12)
V = 0(0)(12)
V = 0
The maximum volume that the box can hold is 0 cubic inches, which means that it cannot hold any volume. This suggests that there might be an error in the problem statement or assumptions.