cylindrical box will be tied up with ribbon
as shown in the figure. The longest piece of ribbon available
is 130 cm long, and 10 cm of that are required for the bow.
Find the radius and height of the box with the largest possible
volume.
Well, well, well, we have a mystery to solve with our ribbon-wrapped cylindrical box! Let's get cracking, shall we?
Now, to find the radius and height that will give us the largest possible volume, we need to put on our detective hats and start investigating.
First, let's figure out the length of ribbon needed to go around the box. We know that the longest piece of ribbon available is 130 cm, and 10 cm is already spoken for by the bow. That means we have 130 - 10 = 120 cm left for the circumference of the box.
The circumference of a cylindrical box is given by the formula 2πr, where r is the radius. So we can set up the equation 2πr = 120 and solve for r. Dividing both sides by 2π, we get r = 120 / (2π) ≈ 19.1 cm.
Next, let's calculate the height of the box. We want to maximize the volume, and the formula for the volume of a cylinder is V = πr²h. Since we know r is approximately 19.1 cm, we need to find the height that will give us the maximum volume.
Now, my comedic intuition suggests that the maximum volume will occur when the height is equal to the radius. It's like a perfect comedy duo, you know? So let's set h = 19.1 cm.
Plugging the values of r = 19.1 cm and h = 19.1 cm into the volume formula, we get V = π(19.1)²(19.1) ≈ 21,898.8 cm³.
So, dear detective, it seems that the radius of the cylindrical box is approximately 19.1 cm and the height is also approximately 19.1 cm. And with these dimensions, the box will have a maximum volume of approximately 21,898.8 cm³. Case closed!
To find the radius and height of the box with the largest possible volume, we can use the calculus method of finding the maximum of a function.
1. Let's start by defining the variables:
- r: radius of the cylinder
- h: height of the cylinder
2. The volume of a cylindrical box is given by the formula:
V = π * r^2 * h
3. We have a constraint that the piece of ribbon available is 130 cm long, and 10 cm are required for the bow. This constraint can be expressed as:
2πr + 2h = 130 - 10
4. Rearrange the constraint equation to solve for h:
2h = 120 - 2πr
h = (120 - 2πr) / 2
h = 60 - πr
5. Substitute the expression for h in terms of r into the volume formula:
V = π * r^2 * (60 - πr)
6. Expand the equation:
V = 60πr^2 - π^2r^3
7. To find the maximum volume, we need to find the critical points of the volume function. We can do this by taking the derivative of V with respect to r and setting it equal to zero:
dV/dr = 120πr - 3π^2r^2 = 0
8. Solve for r:
120πr - 3π^2r^2 = 0
r(120π - 3π^2r) = 0
r = 0 (not a valid solution)
or
120π - 3π^2r = 0
r = (120π) / (3π^2)
r = 40 / π
9. Substitute the value of r back into the expression for h to find the corresponding height:
h = 60 - πr
h = 60 - (40 / π) ≈ 60 - 12.73 ≈ 47.27
10. Therefore, the radius of the box with the largest possible volume is approximately 40 / π cm, and the corresponding height is approximately 47.27 cm.
To find the radius and height of the box with the largest possible volume, we can use calculus and optimization techniques. The volume of a cylindrical box is given by V = πr²h, where V is the volume, r is the radius, and h is the height.
Let's denote the radius of the box as r and the height as h. Since the box is tied up with a ribbon, the length of the ribbon required will be the sum of the height, the circumference of the top of the box, and the circumference of the bottom of the box.
The circumference of a circle is given by C = 2πr. So, the length of the ribbon required is L = h + 2πr + 2πr.
Given that the longest piece of ribbon available is 130 cm long, we have the equation:
L = h + 4πr = 130
We also know that 10 cm of the ribbon is required for the bow, so we can rewrite the equation as:
L = h + 4πr = 130 - 10 = 120
Now, we need to find the maximum volume of the cylindrical box. To do this, we need to maximize the volume function V = πr²h while satisfying the constraint equation L = h + 4πr = 120.
To eliminate h from the equations, we can solve the constraint equation for h:
h = 120 - 4πr
Substituting this value of h into the volume equation, we get:
V = πr²(120 - 4πr)
To find the maximum volume, we need to take the derivative of V with respect to r, set it equal to zero, and solve for r:
dV/dr = 2πr(120 - 4πr) + πr²(-4π) = 0
Simplifying the equation:
240πr - 8π²r² - 4π²r² = 0
Combining like terms:
240πr - 12π²r² = 0
Factoring out the common term of 4πr:
4πr(60 - 3πr) = 0
Setting each factor equal to zero:
4πr = 0 or 60 - 3πr = 0
Since r cannot be zero (it represents the radius), we solve the second equation:
60 - 3πr = 0
3πr = 60
r = 20/π
Thus, we have found the value of r that maximizes the volume.
To find the corresponding value of h, we can substitute this value of r back into the constraint equation:
h + 4πr = 120
h = 120 - 4π(20/π)
h = 120 - 80
h = 40
Therefore, the radius of the box that maximizes the volume is 20/π and the height is 40.
assuming the ribbon is tied on top, and holds on the lid, its length will be
2h+2d + 10 = 130
h+2r = 60
So, since v = pi r^2 h,
v = pi r^2 (60-2r)
= 60pi r^2 - 2pi r^3
dv/dr = 120pi r - 6pi r^2
= 6pir(20-r)
dv/dr=0 at r=20
So, the box has
radius = 20
height = 20