Answer: Let x be the side length of each square cut out from the corners.
After cutting out the squares, the dimensions of the resulting rectangle will be:
length = 360 - 2x
width = 240 - 2x
The height of the box will be equal to the side length of the squares cut out, which is x.
The volume of the box is given by the formula:
V = length × width × height
Substituting the expressions for length, width, and height, we get:
V = (360 - 2x)(240 - 2x)x
Expanding this expression, we get:
V = x(86400 - 120x + 4x^2)
To find the maximum volume, we need to find the value of x that maximizes the expression for V.
We can do this by finding the critical points of the function V(x) and then determining whether these points correspond to a maximum or a minimum.
Taking the derivative of V(x) with respect to x, we get:
V'(x) = 86400 - 240x + 8x^2
Setting V'(x) = 0 to find the critical points, we get:
8x^2 - 240x + 86400 = 0
Dividing both sides by 8, we get:
x^2 - 30x + 10800 = 0
Using the quadratic formula to solve for x, we get:
x = (30 ± √(30^2 - 4(1)(10800))) / 2
x = (30 ± 210) / 2
We can discard the negative solution, since x represents a length and therefore must be positive. Thus, we get:
x = (30 + 210) / 2
x = 120
Therefore, the side length of the squares cut out from the corners should be 120 mm in order to maximize the volume of the box.
To find the maximum volume, we can substitute x = 120 into the expression for V:
V = x(86400 - 120x + 4x^2)
V = 120(86400 - 120(120) + 4(120)^2)
V ≈ 27648000 mm^3
Therefore, the maximum volume of the box is approximately 27,648,000 mm^3, and this maximum is achieved when the side length of each square cut out from the corners is 120 mm.
Explanation: