A box lid is to be made from a rectangular piece of cardboard measuring 72 cm by 216 cm. Two equal squares of side x are to be removed from one end, and two equal rectangles are to be removed from the other end so that the tabs can be folded to form a box with a lid. Find x such that the volume of the box is a maximim
wrong
To find the value of x that maximizes the volume of the box, we need to express the volume in terms of x and then differentiate it with respect to x. Then, by setting the derivative equal to zero and solving for x, we can find the value of x that maximizes the volume.
Let's start by visualizing the box and determining the dimensions. The rectangular piece of cardboard has dimensions of 72 cm by 216 cm. We remove two equal squares from one end, which means the length of these squares is x. The remaining piece on that end will have a length of 216 cm - 2x.
Next, we remove two equal rectangles from the other end. The width of each of these rectangles is x, and their lengths have to be equal to the width of the remaining piece of cardboard. So, the lengths of these rectangles will also be 216 cm - 2x.
Now, let us determine the dimensions of the box. The height of the box will be equal to x, and the length and width of the box will be the dimensions of the remaining piece of cardboard after the squares and rectangles have been removed.
Length of the box: 216 cm - 2x
Width of the box: 72 cm - x
Height of the box: x
The volume V of the box is given by multiplying these dimensions:
V = (216 cm - 2x) * (72 cm - x) * x
To proceed further, let's simplify this expression. Expand the equation:
V = x(72 cm - x)(216 cm - 2x)
V = x(15552 cm^2 - 720 cmx - 432x + 2x^2)
V = x(2x^2 - 1152x + 15552 cm^2)
Now we have an equation for the volume of the box in terms of x. We can find the value of x that maximizes the volume by differentiating V with respect to x:
dV/dx = 2x^2 - 1152x + 15552
Set dV/dx equal to zero and solve for x:
2x^2 - 1152x + 15552 = 0
To solve this quadratic equation, we can factor or use the quadratic formula. Factoring may be difficult due to the large coefficients involved, so let's use the quadratic formula:
x = (-(-1152) ± √((-1152)^2 - 4 * 2 * 15552)) / (2 * 2)
Simplifying further:
x = (1152 ± √(1333248 - 124416)) / 4
x = (1152 ± √(1308832)) / 4
x = (1152 ± 1142.91) / 4
x ≈ (1152 + 1142.91) / 4 ≈ 573.455 cm
or
x ≈ (1152 - 1142.91) / 4 ≈ 2.77 cm
The dimensions of the squares cannot be negative, so we discard the solution x = 2.77 cm. Therefore, the value of x that maximizes the volume of the box is approximately x = 573.455 cm.
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.
have not seen that variation of the popular "make a box" question before.
Made a diagram showing the net of the box with a lid.
let the sides of the square to be cut out be x cm
the the width of the box is 72 - 2x
let the length of the box be y cm
According to my sketch,
2x + 2y = 216
x+y = 108
y = 108-x
the height of the box is x
Volume = V = lwh = (72 - 2x)(108-x)(x)
= x(7776 - 288x - 2x^2)
= 7776x - 288x^2 - 2x^3
dV/dx = 7776 - 576x - 6x^2
= 0 for a max of V
x^2 + 96x - 1296 = 0
(x - 12)(x + 108) = 0
x = 12 or x = a negative, which is silly
so x = 12