A rectangular piece of paper with perimeter
100 cm is to be rolled to form a cylindrical
tube. Find the dimensions of the paper that
will produce a tube with maximum volume.
I have made it up to getting an equation for V(w).
Well, well, well! It looks like we have a rectangular paper going for a roller coaster ride! Let's crunch some numbers and find that mind-boggling maximum volume!
So here's the deal: we need to find the dimensions of the paper that will give us the greatest volume for our cylindrical tube. Let's call the width of the paper "w" and the length "l."
The perimeter of the paper is 100 cm, which means we have a little math equation to play with:
2w + 2l = 100
Now, we need to express the volume of the cylindrical tube in terms of "w" alone. Here comes the fun part!
The formula for the volume of a cylinder is V = πr²h. Since our cylinder is formed from rolling the rectangular paper, the height (h) of the cylinder will be the length of the paper, which is "l."
Now, to find the radius (r) of the cylinder, let's break it down: The circumference of the base of the cylinder will be equal to the perimeter of the paper, which is 100 cm. So we have:
2πr = 100
Solving for r, we get r = 50/π.
Now, the width of the paper, which is "w," will be the length of the base of the cylinder. So we have:
2r + w = l
We can simplify that as l = 2r + w.
Now, let's rewrite the volume of the cylinder in terms of "w" alone. We substitute the values for r and l into the equation:
V(w) = πr²(2r + w)
V(w) = π(50/π)²(2(50/π) + w)
V(w) = (2500/π)(2(50/π) + w)
V(w) = (5000/π²) + (2500/π)w + (2500/π²)w²
So, my dear friend, we've arrived at the equation for the volume of the cylinder, V(w) = (5000/π²) + (2500/π)w + (2500/π²)w². Now, you can work your mathematical magic and find the values of "w" that maximize the volume. Good luck, math guru!
To find the dimensions of the paper that will produce a tube with maximum volume, let's go step by step.
Step 1: Define the variables
Let the width of the rectangular piece of paper be 'w' units (in cm).
Let the length of the rectangular piece of paper be 'l' units (in cm).
Let the radius of the cylindrical tube formed be 'r' units (in cm).
Let the height of the cylindrical tube formed be 'h' units (in cm).
Step 2: Write down the equations
The perimeter of the rectangular piece of paper is given as 100 cm. Since it is a rectangle, we can write the equation for the perimeter as:
2l + 2w = 100 -----------(1)
The volume of the cylindrical tube is given as:
V = πr^2h ---------------(2)
Step 3: Express width in terms of the length
From equation (1), we can express 'l' in terms of 'w' as:
l = (100 - 2w) / 2
l = 50 - w
Step 4: Express height in terms of the radius
The height of the cylindrical tube can be expressed in terms of the radius using geometry. When a rectangular piece of paper is rolled into a cylinder, the height of the cylinder is equal to the width of the paper. Therefore, we can write:
h = w
Step 5: Express volume in one variable
Now, substitute the expressions for 'l' and 'h' from steps 3 and 4 into equation (2) to express the volume 'V' in terms of the width 'w':
V = πr^2h
V = πr^2(w)
Step 6: Express r in terms of w
The circumference of the cylindrical tube is equal to the perimeter of the rectangular piece of paper. Therefore, we can write:
2πr = 2l + 2w
2πr = 2(50 - w) + 2w
2πr = 100 - 2w + 2w
2πr = 100
Divide both sides by 2π to solve for 'r':
r = 100 / (2π)
r = 50 / π
Step 7: Substitute the expression for 'r' in terms of 'w' into the volume equation
Substitute the expression for 'r' from step 6 into equation (2):
V = πr^2w
V = π(50/π)^2w
V = (2500/π)w
Step 8: Simplify the equation for the volume
Simplify the equation for the volume 'V' to get it in its simplest form:
V = (2500/π)w
Step 9: Determine the maximum volume
To find the width 'w' that will give the maximum volume, we need to maximize the equation for 'V'. Since the equation is linear, the maximum volume occurs when 'w' has the largest value possible within the constraints.
As the width 'w' cannot exceed half of the perimeter, which is 50 cm, we can conclude that 'w' should be 50 cm to obtain the maximum volume.
Step 10: Calculate the dimensions
Now that we have determined the width 'w', we can find the corresponding values of 'l', 'r', and 'h'.
Using step 5:
V = (2500/π)w
V = (2500/π)(50)
V ≈ 39788.135
Using step 3:
l = 50 - w
l = 50 - 50
l = 0 cm
Using step 4:
h = w
h = 50 cm
Using step 6:
r = 50 / π
r ≈ 15.92 cm
Therefore, the dimensions of the paper that will produce a tube with maximum volume are:
Width 'w' = 50 cm
Length 'l' = 0 cm
Radius 'r' ≈ 15.92 cm
Height 'h' = 50 cm
Note: The length of the paper becomes zero in order to form the largest possible volume.
To find the dimensions of the paper that will produce a tube with maximum volume, we can follow these steps:
Step 1: Understand the problem
We are given a rectangular piece of paper with a perimeter of 100 cm. We need to roll it to form a cylindrical tube. The goal is to find the dimensions of the paper that will maximize the volume of the tube.
Step 2: Define the variables
Let's define the width of the rectangular piece of paper as w (in cm). Since the length of the paper will be equal to its perimeter divided by twice the width, we can define the length as l = (100 - 2w)/2 = (50 - w) cm.
Step 3: Formulate the equation for the volume of the cylindrical tube
The volume of a cylinder can be calculated using the formula V = πr^2h, where V represents the volume, r is the radius of the base, and h is the height of the cylinder.
In our case, when we roll the rectangular paper, the width becomes the height of the cylindrical tube, and the length becomes the circumference of the base. The radius of the base can be found by dividing the circumference by 2π.
The volume of the cylindrical tube can be expressed as:
V(w) = π * (circumference/2π)^2 * width = π * (l/2π)^2 * w
Step 4: Simplify the equation
Let's simplify the equation by substituting the value of l:
V(w) = π * ((50 - w)/(2π))^2 * w
Step 5: Find the derivative of V(w)
To find the maximum volume, we need to take the derivative of V(w) with respect to w and set it equal to zero.
dV/dw = 2π * (50 - w)/(2π) * (-1/(2π))^2 * w + π * ((50 - w)/(2π))^2 = 0
Simplifying this equation will give us the value of w that maximizes the volume.
I hope this explanation helps you understand the process to get the equation for V(w). If you need further assistance, please let me know.
if the sheet is x by y, and is rolled along the y axis,
2x+2y = 100
v = pi r^2 y
where 2pi r = x, or r = x/(2pi), so
v = pi (x/(2pi))^2 (100-2x)/2
= x^2(50-x)/(4pi)
dv/dx = x(100-3x)/(4pi)
dv/dx=0 when x = 100/3
at that point, max v is pi*(50/3)^3