Question

can company makes a cylindrical can that has a radius of 6 cm and a height of 10 cm. One of the company's clients needs a cylindrical can that has the same volume but is 15 cm tall. What must the new radius be to meet the client's need? Round to the nearest tenth of a centimeter.

Answer 1

the new radius be to meet the client's need is 4.9 cm .

Step-by-step explanation:

Here we have , can company makes a cylindrical can that has a radius of 6 cm and a height of 10 cm. One of the company's clients needs a cylindrical can that has the same volume but is 15 cm tall. We need to find What must the new radius be to meet the client's need . Let's find out:

Let we have two cylinders of volume $V_1 , V_2$ with parameters as follows :

$r_1=6cm\\h_1=10cm\\r_2=?\\h_2=15cm$

We know that volume of cylinder is $\pi r^2h$ , According to question volume of both cylinder is equal i.e

⇒ $V_1=V_2$

⇒ $\pi (r_1)^2h_1= \pi (r_2)^2h_2$

⇒ $(r_1)^2h_1= (r_2)^2h_2$

⇒ $\frac{(r_1)^2h_1}{h_2}= (r_2)^2$

⇒ $(r_2) =\sqrt{ \frac{(r_1)^2h_1}{h_2}}$                   Putting all values

⇒ $(r_2) =\sqrt{ \frac{(6)^2(10)}{15}}$

⇒ $(r_2) =\sqrt{ \frac{36(10)}{15}}$

⇒ $(r_2) =\sqrt{ \frac{360}{15}}$

⇒ $(r_2) =\sqrt{24}$

⇒ $(r_2) =4.9cm$

Therefore , the new radius be to meet the client's need is 4.9 cm .

Answer 2

Answer: Its 4.9 cm

Step-by-step explanation: