Answer:
Explanation:
To find the volume of the box after cutting out squares of length x from each corner, we need to consider the dimensions of the resulting box. After cutting out squares of length x from each corner, the dimensions of the box will be (24-2x) inches by (24-2x) inches, with a height of x inches.
The volume (V) of the box can be expressed as:
V = (24 - 2x) * (24 - 2x) * x
Now, let's simplify the equation:
V = (576 - 48x - 48x + 4x^2) * x
V = (4x^2 - 96x + 576) * x
V = 4x^3 - 96x^2 + 576x
To find the value of x that yields the maximum volume, we need to find the critical points. To do this, we take the derivative of the volume function with respect to x and set it equal to zero:
dV/dx = 12x^2 - 192x + 576
Now, set dV/dx = 0:
12x^2 - 192x + 576 = 0
Divide the equation by 12 to simplify:
x^2 - 16x + 48 = 0
Now, we can use the quadratic formula to solve for x:
x = (-(-16) ± √((-16)^2 - 4*1*48)) / (2*1)
x = (16 ± √(256 - 192)) / 2
x = (16 ± √64) / 2
x = (16 ± 8) / 2
Now, we have two possible values for x:
1. x = (16 + 8) / 2 = 24 / 2 = 12 inches
2. x = (16 - 8) / 2 = 8 / 2 = 4 inches
We discard the negative value of x since it doesn't make sense in the context of the problem (we can't have negative side lengths). So, the valid value for x that yields the maximum volume is approximately 12 inches.
To graph the function, we can plot the volume (V) as a function of x:
V(x) = 4x^3 - 96x^2 + 576x
The x-axis represents the value of x (side length of the squares cut out), and the y-axis represents the volume of the box. The maximum point on the graph corresponds to the value of x that gives the maximum volume, which we have already determined to be approximately 12 inches.