# 2 tricky optimization problems

• Jun 12th 2009, 12:51 PM
Starmix
2 tricky optimization problems
Hey guys,
I'm stuck with these two problems, I tried various ways of solving, but with no luck.

1. A rectangular page is to contain 180 square inches of print. The top and bottom margins are each 0.7 inches wide , and the margins on each side is 2 inches wide. What should the dimensions be if the least amount of material is to be used ?

Length of top(bottom) :
Length of side :

2. Suppose postal requirements are that the maximum of the length plus the girth (cross sectional perimeter) of a rectangular package that may be sent is 250 inches. Find the dimensions of the package with square ends whose volume is to be maximum.

Square side :
Length :

What to do? Thanks.

PS: solutions are due 8 pm today june 12th!
• Jun 12th 2009, 03:05 PM
Soroban
Hello, Starmix!

Who made up these problems?
The answers are quite ugly.

Quote:

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: .$\displaystyle y + 1.4 \:=\:3\sqrt{7} + 1.4$

• Jun 12th 2009, 10:34 PM
Starmix
Merci.....! :))

Actually I did everything the same except for the margins...I subtracted them from x and y values instead of adding.
• Jun 13th 2009, 03:38 AM
Soroban
Hello again, Starmix!

Quote:

Actually I did everything the same except for the margins.
I subtracted them from x and y values instead of adding.

I did, too . . . but the algebra/arithmetic got quite messy.

• Jun 13th 2009, 04:10 AM
Soroban
Hello, Starmix

A bit late, but here's the other one . . .

Quote:

2. Suppose postal requirements are that the maximum of the length plus the girth
(cross sectional perimeter) of a rectangular package that may be sent is 250 inches.
Find the dimensions of the package with square ends whose volume is to be maximum.

Code:

            * - - - *           /      /|           /      / |         /      /  |         /      /  *       /      /  /       * - - - *  /       |      |  / y     x |      | /       |      |/       * - - - *           x

The girth is $\displaystyle 4x$, the length is $\displaystyle y.$
. . $\displaystyle 4x + y \:=\:250 \quad\Rightarrow\quad y \:=\:250-4x$ .[1]

The volume of the box is: .$\displaystyle V \;=\;x^2y$ .[2]

Substitute [1] into [2]: .$\displaystyle V \;=\;x^2(250-4x) \;=\;250x^2 - 4x^3$

Then: .$\displaystyle V' \;=\;500x - 12x^2 \:=\:0 \quad\Rightarrow\quad 4x(125 - 3x) \:=\:0$

And we have: .$\displaystyle \begin{Bmatrix}x \:=\:0 & \text{min.volime} \\ x \:=\:\frac{125}{3} & \text{max.volume} & {\color{red}*}\end{Bmatrix}$

Substitute into [1]: .$\displaystyle y \:=\:250 - 4\left(\tfrac{125}{3}\right) \:=\:\tfrac{250}{3}$

Square side: .$\displaystyle 41\tfrac{2}{3}$ inches.

. . . .Length: .$\displaystyle 83\tfrac{1}{3}$ inches.

• Jun 13th 2009, 04:23 AM
mr fantastic
Quote:

Originally Posted by Starmix
Hey guys,
I'm stuck with these two problems, I tried various ways of solving, but with no luck.

1. A rectangular page is to contain 180 square inches of print. The top and bottom margins are each 0.7 inches wide , and the margins on each side is 2 inches wide. What should the dimensions be if the least amount of material is to be used ?

Length of top(bottom) :
Length of side :

2. Suppose postal requirements are that the maximum of the length plus the girth (cross sectional perimeter) of a rectangular package that may be sent is 250 inches. Find the dimensions of the package with square ends whose volume is to be maximum.

Square side :
Length :

What to do? Thanks.

PS: solutions are due 8 pm today june 12th!

(My highlighting) I assume your teacher has no problem with you getting people to do your homework for you.