using Pi in a 4 inch cylinder A=2Pir^2+2PiRH

Question

A cylinder (round can) has a circular base and a circular top with vertical sides in between. Let r be the radius of the top of the can and let h be the height. The surface area of the cylinder, A, is A=2πr^2+2πrh(two circles, one for the top and one for the bottom plus a rolled-up rectangle for the sides). 

Part a: Assume that the height of your cylinder is 44 inches. Consider AA as a function of rr, so we can write that as A(r)=2πr^2+8πr. What is the domain of A(r)r? In other words, for which values of r is A(r)r defined?

r(A)=

Part c: If the surface area is 200 square inches, then what is the radius r? In other words, evaluate r(200). Round your answer to 2 decimal places.

The radius is _____ inches if the surface area is 200 square inches.

 

Answer 1

Part A. The domain of A(r) is all possible values of r. A radius cannot be negative and it cannot be zero. So the domain is 0 < r < ∞. BTW, I think you mean that the height is h = 4 inches (not 44).

Part B. What is r(A)? r(A) is the inverse of A(r) = 2πr² + 8πr .

To find the inverse, we need to get the variable r by itself on one side of the equation. But there are two r's in the equation. What do you do? Many students get stuck here.

You can do this by completing the square. First, A(r) = 2π [r² + 4r] to make it easy. Then complete the square: A(r) = 2π [r² + 4r + 4 - 4] = 2π [(r + 2)² - 4] = 2π(r + 2)² - 8π. Now there is only one r, so we can get the inverse.

A = 2π(r + 2)² - 8π

A + 8π = 2π(r + 2)²

(A + 8π) / 2π = (r + 2)²

±sqrt((A + 8π) / 2π) = r + 2

±sqrt((A + 8π) / 2π) - 2 = r

Because this is an area, you only need to positive.

So, the answer is: r(A) = sqrt((A + 8π) / 2π) - 2

Part C.

For this part, you can just plug in A = 200

sqrt ((200 + 8π) / 2π) - 2 = 3.99 inches. This is the answer.

To check, you can plug 3.99 into the original equation A(r) = 2πr² + 8πr and get just about 200..