What is the largest possible volume for this cylinder?

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Homework Help Overview

The problem involves a rectangle with a perimeter of 40 cm that is rotated around one of its sides to form a right cylinder. The goal is to determine the largest possible volume for this cylinder, utilizing the relationship between the dimensions of the rectangle and the volume formula for a cylinder.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss how to express the dimensions of the rectangle in terms of the radius and height of the cylinder. There are attempts to relate the perimeter of the rectangle to the volume of the cylinder, with some confusion regarding the definitions of the rectangle's sides and their relationship to the cylinder's dimensions.

Discussion Status

Participants are exploring various interpretations of the problem, particularly how the rectangle's dimensions relate to the cylinder's volume. Some have provided guidance on expressing height in terms of radius and have suggested methods for finding the maximum volume. There is an ongoing clarification regarding the geometric relationships involved.

Contextual Notes

There is some confusion about the relationship between the rectangle and the cylinder, particularly regarding whether the width of the rectangle corresponds to the diameter or radius of the cylinder. The problem states that the rectangle rotates around one of its sides, which influences the interpretation of the dimensions.

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Homework Statement


A rectangle with a perimeter of 40 cm is rotated around one of its sides creating a right cylinder. What is the largest possible volume for this cylinder?

Homework Equations


Volume of a cylinder= pi*r^2*h
Perimeter of a rectangle= 2x + 2L

The Attempt at a Solution


I know one of my equations is
40=2x+2h
And then I isolated h by itself and got:
h= 40-2x / 2
But for my volume equation I have two variables on one side(r & h) and that's where I'm stuck. I don't understand how the perimeter of the rectangle can relate to the volume of the cylinder. How am I suppose to write x in terms of radius? The answer on the back of my textbook says 3723.37cm^3 for what its worth.
 
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abizan said:
How am I suppose to write x in terms of radius?
Draw a picture. How is the radius related to the rectangle?
 
fresh_42 said:
Draw a picture. How is the radius related to the rectangle?
Is the circumference of the circles on the cylinder equal to the lengths of the rectangle? Because I tried that and I did not get the right answer...
 
abizan said:

Homework Statement


A rectangle with a perimeter of 40 cm is rotated around one of its sides creating a right cylinder. What is the largest possible volume for this cylinder?

Homework Equations


Volume of a cylinder= pi*r^2*h
Perimeter of a rectangle= 2x + 2L

The Attempt at a Solution


I know one of my equations is
40=2x+2h
And then I isolated h by itself and got:
h= 40-2x / 2
But for my volume equation I have two variables on one side(r & h) and that's where I'm stuck. I don't understand how the perimeter of the rectangle can relate to the volume of the cylinder. How am I suppose to write x in terms of radius? The answer on the back of my textbook says 3723.37cm^3 for what its worth.

You have used L and h for the same thing. You have one side of the rectangle is x and the other is (40-2x)/2 = 20-x. (Note the necessary parentheses). What is the volume if that rectangle with sides x and 20-x is rotated about an edge?
 
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Imagine a rectangle, say with a long and a short side. Then fix a long stick along the long side. This gives you a flag. Wave it so it circles around your stick. So radius, height and rectangle sides are actually only 2 lengths. Using the perimeter allows you to reduce it to only one left: x.
 
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cylinder1.PNG
Here you see the direct relationship. You have that 40=2(r+h) directly is your 40=2(X+L)=2(r+h).You can express h=20-r. Plug this into your equation of volume, you will get a function V which depends on r, V(r). Now find the maxima, or if you like, the derivative of V with respect to r, that's the slope of the function, now maximum will be if the slope is zero. So \frac {dV}{dr} = 0 You will get an quadratic equation with two solutions, one of them will make sense, and when you get the r_{max} plug that into the volume equation and you will get 3723,37. Try it.
 
Last edited:
Alexiy said:
cylinder1.PNG
Here you see the direct relationship. You have that 40=2(r+h) directly is your 40=2(X+L)=2(r+h).You can express h=20-r. Plug this into your equation of volume, you will get a function V which depends on r, V(r). Now find the maxima, or if you like, the derivative of V with respect to r, that's the slope of the function, now maximum will be if the slope is zero. So \frac {dV}{dr} = 0 You will get an quadratic equation with two solutions, one of them will make sense, and when you get the r_{max} plug that into the volume equation and you will get 3723,37. Try it.
hmm okay I get this but I'm confused as to why the rectangle is only half of the cylinder? If a rectangle forms the cylinder, shouldn't the width of the rectangle be the diameter of the cylinder rather than the radius?
 
abizan said:
hmm okay I get this but I'm confused as to why the rectangle is only half of the cylinder? If a rectangle forms the cylinder, shouldn't the width of the rectangle be the diameter of the cylinder rather than the radius?
Because that's what it says in the text of the problem. The problems states that the rectangle rotates on one of its sides? here's a picture:
cylinder1.png
 
Alexiy said:
Because that's what it says in the text of the problem. The problems states that the rectangle rotates on one of its sides? here's a picture:
cylinder1.png
wow I completely missed that. Thank you so much!
 
  • #10
fresh_42 said:
Imagine a rectangle, say with a long and a short side. Then fix a long stick along the long side. This gives you a flag. Wave it so it circles around your stick. So radius, height and rectangle sides are actually only 2 lengths. Using the perimeter allows you to reduce it to only one left: x.
Oh I get it now so the radius is one of the rectangle's sides. Thanks for the help!
 

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