# Thread: 2 tricky optimization problems

1. ## 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!

2. Hello, Starmix!

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

1. A rectangular page is to contain 180 in² 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 $x$.
The height of the print is $y$.
The area of the print is 180 in²: . $xy = 180 \quad\Rightarrow\quad y = \tfrac{180}{x}$ .[1]

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

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

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

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

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

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

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

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

3. Merci.....! )

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

4. Hello again, Starmix!

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.

5. Hello, Starmix

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

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

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

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

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

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

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

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

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

6. 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.