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What is the largest possible volume for this cylinder?

  1. Jan 16, 2016 #1
    1. The problem statement, all variables and given/known data
    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?

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

    3. The attempt at a solution
    I know one of my equations is
    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 thats 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.
  2. jcsd
  3. Jan 16, 2016 #2


    Staff: Mentor

    Draw a picture. How is the radius related to the rectangle?
  4. Jan 16, 2016 #3
    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...
  5. Jan 16, 2016 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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?
  6. Jan 16, 2016 #5


    Staff: Mentor

    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.
  7. Jan 16, 2016 #6
    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, thats the slope of the function, now maximum will be if the slope is zero. So [itex] \frac {dV}{dr} = 0 [/itex] You will get an quadratic equation with two solutions, one of them will make sense, and when you get the [itex] r_{max} [/itex] plug that into the volume equation and you will get 3723,37. Try it.
    Last edited: Jan 16, 2016
  8. Jan 16, 2016 #7
    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?
  9. Jan 16, 2016 #8
    Because thats what it says in the text of the problem. The problems states that the rectangle rotates on one of its sides? heres a picture:
  10. Jan 16, 2016 #9
    wow I completely missed that. Thank you so much!!
  11. Jan 16, 2016 #10
    Oh I get it now so the radius is one of the rectangle's sides. Thanks for the help!
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