What is the maximum volume of a right circular cylinder with surface area 24𝜋 centimeters squared? Give your answer in terms of 𝜋.
This problem is about a right circular cylinder, so let’s draw one. And what are we told about this right circular cylinder? The given information is that it has a surface area of 24𝜋 centimeters squared. But there are many right circular cylinders with a surface area of 24𝜋 centimeters squared. Some will be long and thin, and some will be a short and wide. Our task is to find the maximum volume that such a cylinder can have.
This is an optimization problem that you can imagine that turning up in real life, you’re given a certain amount of sheet metal, 24𝜋 centimeters squared to be precise, and you have to make the biggest tin can possible using that sheet metal. Let’s start by introducing some variables. We’ll call the radius of the right circular cylinder 𝑟. And the height of the right circular cylinder we’ll call ℎ. The values of 𝑟 and ℎ completely determine the size and shape of our right circular cylinder. And we can write the surface area and volume of our cylinder in terms of 𝑟 and ℎ.
The surface area of the cylinder is two 𝜋𝑟 squared plus two 𝜋𝑟ℎ. The two 𝜋𝑟 squared comes from the two plane faces of our cylinder, the top and bottom of the tin can if you will, which are both discs of radius 𝑟. The two 𝜋𝑟ℎ is the remaining curved surface area of the can, where the label would go. You can imagine that taking off this label and flattening it out into a rectangle of width two 𝜋𝑟 and height ℎ, which explains why the curved surface area is two 𝜋𝑟 times ℎ.
We also have a formula for the volume of a right circular cylinder with radius 𝑟 and height ℎ: it’s the area of the base of the right circular cylinder 𝜋𝑟 squared times the height of the cylinder ℎ. Remember that the volume 𝑉 is a quantity that we want to maximize. If we had of 𝑉 as a function of some variable, we would be able to use calculus techniques to find the maximum of that function over a given interval.
Unfortunately we don’t have a 𝑉 in terms of a single variable which would make it function of that variable. We have it in terms of both 𝑟 and ℎ. But there’s some information we haven’t used yet: the surface area of the right circular cylinder is 24𝜋 centimeters squared. Using the expression for the surface area 𝐴 in terms of 𝑟 and ℎ, we find working in units of centimeters that two 𝜋𝑟 squared plus two 𝜋𝑟ℎ is 24𝜋.
We can divide through by two 𝜋 on both sides to find that 𝑟 squared plus 𝑟ℎ is 12. Okay, we found a relation between the variables 𝑟 and ℎ. But how does this help us maximize the volume 𝑉 equals 𝜋𝑟 squared ℎ, which is after all what we’re trying to do? Well, we can use this relation to write ℎ in terms of 𝑟. And substituting this expression for ℎ, we’ll get 𝑉 in terms of 𝑟 alone, 𝑉 as a function of 𝑟, which we then know how to maximize.
So let’s do this. We subtract 𝑟 squared from both sides. And dividing through by 𝑟, we find ℎ in terms of 𝑟: ℎ equals 12 minus 𝑟 squared over 𝑟. We substitute this expression for ℎ in terms of 𝑟 into the formula for the volume of 𝑉. We now have 𝑉 as a function of 𝑟: 𝑉 equals 𝜋𝑟 squared times 12 minus 𝑟 squared over 𝑟. And we can simplify somewhat to find that 𝑉 is 𝜋𝑟 times 12 minus 𝑟 squared.
Before we worry about maximizing 𝑉, let’s first think about what this relation means. This is the formula for the volume of a right circular cylinder with a surface area 24𝜋. Given any right circular cylinder with surface area 24𝜋, you can substitute its radius for 𝑟 in this formula to get its volume. Okay, so we have 𝑉 as a function of 𝑟, and we want to maximize 𝑉. It’s calculus time! But hang on, over what interval of 𝑟 are we maximizing 𝑉?
Remember that 𝑟 is the radius of our right circular cylinder. And as a radius, it must be positive. Is there an upper bound for 𝑟? Well, this is less obvious. But you might notice that if 𝑟 is greater than the square root of 12, then 12 minus 𝑟 squared, and hence the volume, will be negative. And clearly this doesn’t make sense. What’s going on here? Well if 𝑟 is greater than the square root of 12, then the area of just the two plane faces of the right circular cylinder, two 𝜋𝑟 squared, is greater than 24𝜋.
As the total surface area of the cylinder is 24𝜋 centimeters squared, the area of the plane faces must certainly be less than 24𝜋 centimeters squared. And so 𝑟 must be less than root 12 centimeters, as the curved surface area can’t be negative. Okay, so what is the interval? 𝑟 has to be between zero and the square root of 12. And I suppose we can also consider cylinders of radius zero, although they don’t seem very cylinder-y to me because it makes our interval closed.
We’ve managed to turn this geometry problem into a problem about maximizing a function over a closed interval. Now we can apply the closed interval method. To find the absolute maximum and minimum values of a function on an interval, you find the values of the function at the critical numbers and at the intervals’ endpoints. The largest of these values is the absolute maximum on interval, and the smallest of the values is the absolute minimum.
Let’s clear some space. First, let’s find the critical numbers of the function, if there are any. The critical numbers are numbers for which the derivative of the function 𝑉 with respect to 𝑟 is zero or undefined. So we have to differentiate 𝑉 with respect to 𝑟. We use our expression for 𝑉 in terms of 𝑟. We expand this expression to make it easier to differentiate. Here, we use the formula for the derivative of a number times a power of a variable with respect to that variable. And this formula works just as well for 𝑟 as it does for 𝑥.
Differentiating term by term then, we get 12𝜋 minus three 𝜋𝑟 squared. And remember, we’re looking for the critical numbers; that is, values of 𝑟 for which the derivative is zero or doesn’t exist. Clearly, the derivative exists for any real number 𝑟. It’s just a quadratic in 𝑟. And so we only have to worry about finding the values of 𝑟 for which the derivative is zero. Taking out a common factor of three 𝜋, we see that the derivative is three 𝜋 times four minus 𝑟 squared. We factor the difference of two squares into two minus 𝑟 times two plus 𝑟. And hence, we see that the critical numbers are two and negative two.
Okay, let’s clear some space. We’ve found the critical numbers then, and we know what the interval endpoints are. The endpoints are zero and root 12. We notice then that one of our critical numbers, negative two, is not in our interval. And so we don’t need to consider it. Remember that the closed interval method tells us that one of these numbers — two, zero, or the square root of 12 — is where the absolute maximum of 𝑉 is attained. And also one of them is where the absolute minimum value of 𝑉 on the interval is attained, although we care about this less.
We now just need to evaluate 𝑉 for each of these values of 𝑟. Let’s start with two. When 𝑟 is two, what is 𝑉? We substitute two for 𝑟 in our expression for 𝑉 to get 𝜋 times two times 12 minus two squared. Simplifying and writing the units back in, remember we’re working in centimeters and so the volume has units of the centimeters cubed, we get 16𝜋 centimeters cubed when 𝑟 equals two.
How about for 𝑟 equals zero? The volume 𝑉 is 𝜋 times a zero times 12 minus zero squared, which adding the units simplifies to zero centimeters cubed. This isn’t at all surprising: of course a right circular cylinder with radius zero is going to have a volume of zero! Now how about for the other endpoint, root 12? We get 𝜋 times root 12 times 12 minus root 12 squared. We have a zero inside the parentheses, and so the volume is zero here too.
If you work it out, having a radius of root 12 centimeters forces the right circular cylinder to have a height of zero centimeters. And so again it’s not surprising that the volume is zero. We have three volumes 𝑉 for different right circular cylinders of surface area 24𝜋 centimeters squared. And the closed interval method tells us that the largest of these values is the absolute maximum on the interval. That’s the value that we’re looking for, the maximum volume, 𝑉.
Clearly, the largest of these is 16𝜋 centimeters cubed. This is the absolute maximum of the function 𝑉 of 𝑟 on the interval. And so it is the maximum volume of a right circular cylinder with the surface area 24𝜋 centimeters squared. Let’s quickly recap how we solved this problem. We noticed that this problem was about a right circular cylinder so we drew one, and we laid down variables 𝑟 and ℎ. We found that the given information was that the surface area of this cylinder was 24𝜋 centimeters squared, and we were required to find the maximum volume.
And we expressed these quantities in terms of the variables that we laid down, 𝑟 and ℎ. Although we initially had the volume 𝑉 that we wanted to maximize in terms of both 𝑟 and ℎ, we used the given information about the surface area of the cylinder to write a relation between 𝑟 and ℎ, hence writing ℎ in terms of 𝑟 and then 𝑉 in terms of 𝑟 alone. We then had a calculus problem: we wanted to maximize 𝑉, a function of 𝑟, over a certain interval, which we found again considering the context.
We applied the closed interval method, finding the critical numbers of the function 𝑉 of 𝑟 by differentiating and then evaluating 𝑉 at the critical number found as well as the endpoints of the interval. The largest of these values, 16𝜋 centimeters cubed, is the absolute maximum of the function 𝑉 of 𝑟 on the interval of 𝑟. That was relevant. And so it is the maximum volume of a right circular cylinder with surface area 24𝜋 centimeters squared.