Question

You are asked to build an open cylindrical can (i.e. no top) that will hold 364.5 cubic inches. To do this, you will cut its bottom from a square of metal and form its curved side by bending a rectangular sheet of metal.



(a) Express the total amount of material required for the square and the rectangle in terms of r.

Answer 1

Answer:

Formula for the volume of the cylinder is given by;

$V = \pi r^2h$

where r is the radius of the cylinder and h be the height of the cylinder.

As per the statement:

Volume of cylinder(V) = 364.5 cubic inches.

You will cut its bottom from a square of metal and form its curved side by bending a rectangular sheet of metal.

Diameter of a circle = 2r

Then, side of the square (s) = 2r

Area of a square = $s^2 = (2r)^2 = 4r^2$ square inches.

Since, Volume of the cylinder(V) = $\pi r^2h$

then;

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

Substitute the value of V then we have

$h = \frac{364.5}{\pi r^2}$

To find the total amount of material required for the square and the rectangle in terms of r.

$\text{Total amount of material} = \text{area of the square which used to cut the bottom} + \text{curved surface of cylinder}$

Curved surface area of cylinder = $2 \pi rh$

Substitute the value of h we have;

Curved surface area of cylinder = $2 \pi r \cdot \frac{364.5}{\pi r^2}$

                                                      = $\frac{729}{r}$ square inches.

then;

$\text{Total amount of material} = \frac{729}{r}+4r^2$

                                                           = $\frac{4r^3+729}{r}$ square inches

Therefore, the total amount of material required for the square and the rectangle in terms of r is, $\frac{4r^3+729}{r}$ square inches

Answer 2

The total amount of material for the square is $\boxed{4r^{2}}$ and for the rectangle is $\boxed{\dfrac{729}{r}}$.

Further explanation:

The volume of the cylinder is $364.5\text{ in}^{3}$.

The area of the square can be obtained by the formula $\boxed{a^{2}}$ where, $a$ is the side of the square.

The area of the rectangle is obtained by the formula is $\boxed{l\times b}$ , where $l$ and $b$ are length and breadth of the rectangle respectively.

The volume of the cylinder is $\boxed{\pi r^{2}h}$ where, $r$ is the radius of the cylinder and $h$ is the height of the cylinder.

The area of circle is $\boxed{\pi r^{2}}$.

First we have to find the height of the cylinder by applying the formula of volume of cylinder,

$\begin{aligned}\pi r^{2}h&=364.5\\h&=\dfrac{364.5}{\pi r^{2}h}\end{aligned}$  

The base of the cylinder is circle and it is obtained from the square.

Therefore, the diameter of the circle is equal to the side of square.

$\boxed{d=2r=a}$

Here, $a$ is the side of square.

The area of the square is calculated as follows:

$\begin{aligned}\text{Area of square}&=(2r)^{2}\\&=4r^{2}\end{aligned}$

Therefore, the total amount of material require for square is $\boxed{4r^{2}}$.

When we fold the rectangle the length of the rectangle is equal to the perimeter of the base of the cylinder.

$\boxed{2\pi r=l}$

The breadth of the rectangle is equal to the height of the cylinder.

$\boxed{b=h}$  

The value of $h$ is obtained as $\frac{364.5}{r^{2}}$.

Therefore, the area of rectangular sheet of metal is calculated as follows:

$\begin{aligned}\text{Area of rectangle}&=l\times b\\&=(2\pi r)\left(\dfrac{364.5}\pi {r^{2}}\right)\\&=2\left(\dfrac{3645}{10}\right)\left(\dfrac{\pi r}{\pi r^{2}}\right)\\&=\dfrac{7290}{10}\cdot \dfrac{1}{r}\\&=\dfrac{729}{r}\end{aligned}$

 

Thus, the total amount of material require for rectangle is $\boxed{\dfrac{729}{r}}$.

Learn more:

1. What is the general form of the equation of the given circle with center a? https://brainly.com/question/1506955

2. Which equation represents a circle with the same radius as the circle shown but with a center at (-1, 1)? https://brainly.com/question/1952668

Answer details:

Grade: High school

Subject: Mathematics

Chapter: Mensuration

Keywords:  Rectangle, square, cylinder, amount of material, 729/r , 364.5/pir^2 ,area of circle, volume of cylinder,  Material,  cylinderical can, rectangular sheet.