Hello, Starmix!

Who made up these problems?

The answers are quite ugly.

1. A rectangular page is to contain 180 inē of print.

The top and bottom margins are each 0.7 in wide, and the margins on each side is 2 in wide.

What should the dimensions be if the least amount of material is to be used? Did you make a sketch? Code:

: 2 : - x - : 2 :
- *---------------* -
0.7 | | :
- | *-------* | :
: | | | | :
: | | | | :
y | |y | | : y + 1.4
: | | | | :
: | | x | | :
- | *-------* | :
0.7 | | :
- *---------------* -
: - - x + 4 - - :

The width of the print is $\displaystyle x$.

The height of the print is $\displaystyle y$.

The area of the print is 180 inē: .$\displaystyle xy = 180 \quad\Rightarrow\quad y = \tfrac{180}{x}$ .[1]

The width of the page is: $\displaystyle x+4$.

The height of the page is: $\displaystyle y + 1.4$.

The area of the page is: .$\displaystyle A \;=\;(x+4)(y+1.4)$ .[2]

Substitute [1] into [2]: .$\displaystyle A \;=\;(x + 4)\left(\tfrac{180}{x} + 1.4\right)$

. . which simplifies to: .$\displaystyle A \;=\;1.4x + 720x^{-1} + 185.6$

Then we have: .$\displaystyle A' \;=\;1.4 - 720x^{-2} \:=\:0 \quad\Rightarrow\quad 1.4x^2 - 720 \:=\:0 $

. . Then: .$\displaystyle x^2 \:=\:\tfrac{720}{1.4} \:=\:\tfrac{3600}{7} \quad\Rightarrow\quad x \:=\:\tfrac{60}{\sqrt{7}} \:=\:\frac{60\sqrt{7}}{7}$

Substitute into [1]: .$\displaystyle y \:=\:\frac{180}{\frac{60}{\sqrt{7}}} \:=\:3\sqrt{7}$

The width of the page is: .$\displaystyle x + 4 \:=\:\frac{60\sqrt{7}}{7} + 4$

The height of the page is: .$\displaystyle y + 1.4 \:=\:3\sqrt{7} + 1.4$