Question

the diagram shows a cylinder. the original cylinder has a diameter of 31.0 ft and a height of 12.0 ft. the company constructs a new cylindrical container that has the same height as the old container. the volume of the new container is 3 times the volume of the old container. what is the approximate diameter of the new container? a. 26.6 ft b. 36.0 ft c. 49.5 ft d. 51.2 ft e. 93.0 ft

Explanation:

Step1: Recall volume formula for cylinder

$V = \pi r^{2}h$

Step2: Find radius of old - cylinder

Given old - cylinder diameter $d_1=31.0$ ft, so radius $r_1=\frac{d_1}{2}=\frac{31.0}{2}=15.5$ ft, and height $h = 12.0$ ft. Volume of old - cylinder $V_1=\pi r_1^{2}h=\pi\times(15.5)^{2}\times12.0$.

Step3: Set up relationship between volumes

Let the radius of new - cylinder be $r_2$. Since the height of new - cylinder is the same as old one ($h$) and volume of new - cylinder $V_2 = 4V_1$. And $V_2=\pi r_2^{2}h$, $V_1=\pi r_1^{2}h$. So $\pi r_2^{2}h = 4\times\pi r_1^{2}h$.

Step4: Solve for $r_2$

Cancel out $\pi$ and $h$ from both sides of the equation $\pi r_2^{2}h = 4\times\pi r_1^{2}h$, we get $r_2^{2}=4r_1^{2}$. Then $r_2 = 2r_1$. Substitute $r_1 = 15.5$ ft, so $r_2=2\times15.5 = 31$ ft.

Step5: Find diameter of new - cylinder

Diameter $d_2 = 2r_2$. Substitute $r_2 = 31$ ft, we get $d_2=2\times31=57.2$ ft.

Answer:

D. 57.2 ft