# Simple word problem

1. Jan 16, 2016

### abizan

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
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 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. Jan 16, 2016

### Staff: Mentor

Draw a picture. How is the radius related to the rectangle?

3. Jan 16, 2016

### abizan

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...

4. Jan 16, 2016

### LCKurtz

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?

5. Jan 16, 2016

### 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.

6. Jan 16, 2016

### Alexiy

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 $\frac {dV}{dr} = 0$ 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: Jan 16, 2016
7. Jan 16, 2016

### abizan

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?

8. Jan 16, 2016

### Alexiy

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:

9. Jan 16, 2016

### abizan

wow I completely missed that. Thank you so much!!

10. Jan 16, 2016

### abizan

Oh I get it now so the radius is one of the rectangle's sides. Thanks for the help!