The volume of a right circular cylinder is 128𝜋 centimeters cubed. Find the height and base radius of the cylinder such that its surface area is as small as possible.
Let’s start with what we know. We know that the volume of a right circular cylinder equals 𝜋 times the radius squared times the height. 𝜋𝑟 squared is the area of its circular base. And we multiply that by the height of the cylinder. For our particular cylinder, the volume equals 128𝜋 centimeters cubed. And the surface area of a right circular cylinder can be found by multiplying two times 𝜋𝑟 squared plus two 𝜋𝑟 times the height.
We want to optimize the surface area, under the constraints or the conditions that the volume must equal 128𝜋 centimeters cubed. To do that, we wanna take the derivative of the surface area formula. However, right now, our surface area has two variables in the formula, the radius and the height. And we would prefer to just have one variable.
We can use our constraints, the volume and the formula for volume, to rewrite the height in terms of the radius. To do that, we’ll divide both sides of this equation by 𝜋𝑟 squared. On the right side, we’re left with the height. On the left, the 𝜋 in the numerator and the 𝜋 in the denominator cancel out. So we’ll say that the height must be equal to 128 divided by the radius squared. And that means, in place of ℎ, the height, in our surface area formula, we can substitute 128 over 𝑟 squared.
At this point, we notice that there is an 𝑟 in the numerator and an 𝑟 squared in the denominator. And that can simplify so that the formula for the area is equal to two 𝜋𝑟 squared plus two 𝜋 times 128 over 𝑟. We can pull out the factor two 𝜋. To optimize, we first take the derivative of the surface area formula, the derivative of the surface area with respect to 𝑟. And taking the derivative, it’ll be easier if we rewrite 128 over 𝑟 as 128 times 𝑟 to the negative one power.
We know that our constant, two 𝜋, stays the same. The derivative of 𝑟 squared is two times 𝑟 to the first power. Then we’ll take the derivative of 128 times 𝑟 to the negative one power, which is negative 128 times 𝑟 to the negative two power. Like I said earlier, optimizing means taking the derivative of this formula and then setting it equal to zero. Two 𝜋 times two 𝑟 minus 128𝑟 to the negative two power equals zero. We can rewrite negative 128𝑟 to the negative two power as 128 over 𝑟 squared.
And now we need to solve for 𝑟. First, we divide both sides of the equation by two 𝜋. Two 𝜋 divided by two 𝜋 cancels out. And zero divided by two 𝜋 equals zero. On the left, we need to find a common denominator. We can multiply two 𝑟 by 𝑟 squared over 𝑟 squared. That would give us two 𝑟 cubed over 𝑟 squared minus 128 over 𝑟 squared, which we can simplify to say two times 𝑟 cubed minus 128 over 𝑟 squared equals zero. Then multiply both sides of the equation by 𝑟 squared. Two 𝑟 cubed minus 128 equals zero. We add 128 to both sides. Two 𝑟 cubed equals 128. Divide both sides of the equation by two. 𝑟 cubed equals 64. And then take the cube root of 𝑟 cubed and the cubed root of 64, and 𝑟 equals four. The optimized radius equals four.
Our volume was given in centimeters cubed, which tells us that our radius is being measured in centimeters. To find the height, we’ll use the formula that we have that compares the height to the radius. The height equals 128 divided by the radius squared. The height would be 128 divided by four squared. That’s 128 divided by 16, which equals eight. And again, we’ll be measuring in centimeters.
Before we leave this problem, we can go back and plug in the radius and the height to check and make sure that we have gotten the volume we intended. Volume equals 𝜋𝑟 squared times height. 𝜋 times four squared times eight equals 128𝜋, which is what our constraints were. To create a surface area that is as small as possible while maintaining a volume of 128𝜋, a right circular cylinder would need a radius of four centimeters and a height of eight centimeters.