To find the value of x that maximizes the volume of the box, we need to follow these steps:
Step 1: Visualize the box and identify its dimensions.
Let's start by visualizing the box. The given piece of cardboard measures 72 cm by 216 cm. We need to remove two squares of side x from one end of the cardboard and two rectangles from the other end. Let's assume the width of the cardboard is the dimension perpendicular to the side being removed. Therefore, let's call the width of the cardboard w and the length l.
Step 2: Express the dimensions of the box in terms of x.
Since we are removing two squares of side x from one end, the length of the box will be l - 2x. The width of the box will be w, and the height will be x.
Step 3: Write the equation for the volume of the box.
The volume of a rectangular box is calculated by multiplying its length, width, and height. In this case, the volume V can be expressed as:
V = (l - 2x) * w * x
Step 4: Simplify the equation.
Multiply (l - 2x) by w and simplify the equation:
V = lwx - 2wx^2
Step 5: Find the derivative of the volume equation.
To find the value of x that maximizes the volume, we need to find the critical points of the volume equation. We can do this by taking the derivative of the equation with respect to x. Applying the power rule, the derivative dV/dx is:
dV/dx = lw - 4wx
Step 6: Set the derivative equal to zero.
To find the critical points, we need to solve the equation dV/dx = 0. Set the derivative equal to zero and solve for x:
lw - 4wx = 0
Step 7: Solve for x.
Divide both sides of the equation by 4w:
-4wx = -lw
x = lw / 4w
x = l / 4
Step 8: Evaluate the second derivative.
To determine if the value of x we found in step 7 corresponds to a maximum or minimum, we need to evaluate the second derivative of the volume equation. The second derivative d²V/dx² can be found by taking the derivative of the derivative:
d²V/dx² = -4w
Step 9: Determine the nature of the critical point.
Since the second derivative is a constant value (-4w), and it is negative, we can conclude that the critical point is a maximum.
Step 10: Calculate the maximum volume.
Substitute the value of x we found in step 7 back into the volume equation to get the maximum volume:
V = (l - 2x) * w * x
V = (l - 2(l / 4)) * w * (l / 4)
V = (l - l / 2) * w * (l / 4)
V = (l / 2) * w * (l / 4)
V = lw² / 8
So, the maximum volume of the box can be expressed as lw² / 8.