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

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

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

Code:

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

We see that the radius of the cylinder is *half* the side of the square:

. . $\displaystyle r \:=\:\frac{3\sqrt{2}}{2}\text{ inches.}$

The volume of the cylinder is: .$\displaystyle V_2 \:=\:\pi r^2h \:=\:\pi\left(\frac{3\sqrt{2}}{2}\right)^2(36) \:=\:162\pi\text{ in}^2$

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