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Algebra Difficulty 3.6 AMC 10/12 Find the answer Canada

An aluminum can in the shape of a cylinder is closed at both
ends. Its surface area is $300\$300\text{}
cm}^2$. If the radius of the can were doubled, its surface area
would be 900 cm2900\text{ cm}^2. If
instead the height of the can were doubled, what would its surface area
be?

(The surface area of a cylinder with radius rr and height hh is equal to 2πr2+2πrh2\pi r^2 + 2\pi rh.)

Pick one

Solution

Suppose that the original can has radius rr cm and height hh cm.

Since the surface area of the original can is 300 cm2300\text{ cm}^2, then 2πr2+2πrh=3002\pi r^2 + 2\pi r h = 300.

When the radius of the original can is doubled, its new radius is 2r2r cm, and so an expression for its
surface area, in cm2\text{cm}^2, is
2π(2r)2+2π(2r)h2\pi (2r)^2 + 2\pi (2r) h which
equals 8πr2+4πrh8\pi r^2 + 4\pi rh, and so
8πr2+4πrh=9008 \pi r^2 + 4\pi rh = 900.

When the height of the original can is doubled, its new height is 2h2h cm, and so an expression for its
surface area, in cm2\text{cm}^2, is
2πr2+2πr(2h)2\pi r^2 + 2\pi r (2h) which equals
2πr2+4πrh2\pi r^2 + 4\pi rh.

Multiplying $2π\$2\pi r^2 + 2π2\pi r h =
300by3,weobtain by 3, we obtain 6π6\pi r^2 + 6π6\pi
r h = 900$.

Since 8πr2+4πrh=9008 \pi r^2 + 4\pi rh = 900, we
obtain 6πr2+6πrh=8πr2+4πrh2πrh=2πr2πrh=πr2\begin{align*} 6\pi r^2 + 6\pi r h & = 8 \pi r^2 + 4\pi rh \\ 2\pi r h & = 2\pi r^2 \\ \pi r h & = \pi r^2\end{align*} Since 2πr2+2πrh=3002\pi r^2 + 2\pi r h = 300 and πrh=πr2\pi rh = \pi r^2, then 2πr2+2πr2=3002\pi r^2 + 2\pi r^2 = 300 and so 4πr2=3004\pi r^2 = 300 or πr2=75\pi r^2 = 75.

Since πrh=πr2=75\pi rh = \pi r^2 = 75, then
$2π\$2\pi r^2 + 4π4\pi rh = 6 \cdot 75 =
450$, and so the surface area of the cylinder with its height
doubled is 450 cm2450\text{ cm}^2.

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.