Right Circular Cylinder problem...

**orochimaru700** · January 10th 2010, 03:30 AM

1.) Find the waste in making the largest possible cylindrical rod from a bar of iron 3ft long which has a cross section whose diagonal is 6 inches.

2.) A certain factory manufactures tin cans. It received an order for 10,000 cylindrical tin cans of height 5 inches and diameter 3 inches. How many square feet of tin did this order require if 7 sq. in. are allowed for waste and seams in each can.

3.) A circular oak table top is 4ft. in diameter and 3/4 inches thick. How heavy is it if oak weighs 47 lb. per cu. ft.?

**Henryt999** · January 10th 2010, 04:14 AM

Not really sure about the first one but the second one:

Use the Cylinder volume fomula:

Area: $pi*r^2*h$ = $pi*(3/2)^2*5$
Subtract the waste and multiply it by 10000.

Question 3) Again use the area formula: pi*2^2*3/4. and and you will se the volume now subtract by the denisty and there u have it.

**earboth** · January 10th 2010, 04:41 AM

Originally Posted by orochimaru700

1.) Find the waste in making the largest possible cylindrical rod from a bar of iron 3ft long which has a cross section whose diagonal is 6 inches.

2.) A certain factory manufactures tin cans. It received an order for 10,000 cylindrical tin cans of height 5 inches and diameter 3 inches. How many square feet of tin did this order require if 7 sq. in. are allowed for waste and seams in each can.

3.) A circular oak table top is 4ft. in diameter and 3/4 inches thick. How heavy is it if oak weighs 47 lb. per cu. ft.?

To #1) Which shape has the iron bar? (Square, ellipse, triangle, ...?)

to #2) The surface of a cylinder is calculated by:

$a=\underbrace{2 \cdot \pi r^2}_{\text{lid and base}}+\underbrace{2 \pi r \cdot h}_{\text{curved surface}}$

To this value you have to add the waste to get the total amount of material to produce one can.

to #3) The weight of a solid with uniform density is calculated by:

$weight = volume \cdot density$

The problem here is that you have to change all different units into one:

I would use only inches as unit for length:

4 feet ---> ??? inches
1 cubic foot ----> ???? cubic inches

and so on.

**Henryt999** · January 10th 2010, 04:45 AM

I´m sorry but didn´t the second question involve the volume of the cylinder instead of the area?

**orochimaru700** · January 10th 2010, 04:51 AM

@ earboth, don't know what kind of iron bar, this questions were based on Kern and Bland Solid Mensuration with proofs... this is our homework, and seems the toughest, whoever get's the perfect solution and answer in our class get's the honor's rating instantly! Thanks for the tip, i'll try my best

**earboth** · January 10th 2010, 05:00 AM

I´m sorry but didn´t the second question involve the volume of the cylinder instead of the area?

Originally Posted by orochimaru700

... How many square feet of tin did this order require if 7 sq. in. are allowed for waste and seams in each can.

...

According to the text the question asks an area, not a volume.

**Soroban** · January 10th 2010, 06:32 AM

Hello, orochimaru700!

1) Find the waste in making the largest possible cylindrical rod
from a bar of iron 3ft long which has a cross section whose diagonal is 6 inches.

I will assume that the bar has a square cross-section.

Code:

      *-----------*
      |         * |
      |    6  *   |
      |     *     | x
      |   *       |
      | *         |
      *-----------*
            x

The diagonal is 6 inches.

We have: . $x^2+x^2\:=\:6^2 \quad\Rightarrow\quad 2x^2 \:=\:36 \quad\Rightarrow\quad x^2 \:=\:\sqrt{18}$

The side of the square is: . $x\:=\:3\sqrt{2}$ inches.

The volume of the bar is: . $V_1 \:=\:L\cdot W\cdot H \:=\:(3\sqrt{2})(3\sqrt{2})(36) \:=\:648\text{ in}^3$

Code:

                 _
      : - - -  3√2  - - - :
      *-------*-*-*-------*
      |   *           *   |
      | *               * |
      |*                 *|
      |                   |
      *                   *
      *         * - - - - *
      *              r    *
      |                   |
      |*                 *|
      | *               * |
      |   *           *   |
      *-------*-*-*-------*
                :  - r -  :

We see that the radius of the cylinder is half the side of the square:
. . $r \:=\:\frac{3\sqrt{2}}{2}\text{ inches.}$
The volume of the cylinder is: . $V_2 \:=\:\pi r^2h \:=\:\pi\left(\frac{3\sqrt{2}}{2}\right)^2(36) \:=\:162\pi\text{ in}^2$

The waste is: . $V_1 - V_2 \;=\;648 - 162\pi \;\approx\;139.062\text{ in}^3$

**Soroban** · January 10th 2010, 07:18 AM

Hello again, orochimaru700!

2) A certain factory manufactures tin cans.
It received an order for 10,000 cylindrical tin cans of height 5 inches and diameter 3 inches.
How many square feet of tin did this order require
if 7 sq. in. are allowed for waste and seams in each can?

earboth is correct!

The top and bottom require: . $2 \times \pi\left(\frac{3}{2}\right)^2 \:=\:\frac{9\pi}{2}\text{ in}^2$

The side is a rectangle: . $3\pi\text{ inches wide and }5\text{ inches high}\:=\:15\pi\text{ in}^2.$

Each can requires: . $\frac{9\pi}{2} + 15\pi \:=\:\frac{39\pi}{2}\text{ in}^2\text{ of tin.}$

$7\text{ in}^2$ is allowed for waste.
. . So each can is allowed: . $\frac{39\pi}{2} + 7 \:=\:\frac{53\pi}{2}\text{ in}^2\text{ of tin.}$

Ten thousand cans require: . $10,\!000 \times \frac{53\pi}{2} \:=\:265,\!000\pi \text{ in}^2\text{ of tin,}$

. . or: . $\frac{265,000\pi}{144} \;\approx\;5781.4\text{ ft}^2\text{ of tin.}$

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