2 tricky optimization problems

**Starmix** · June 12th 2009, 12:51 PM

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!

**Soroban** · June 12th 2009, 03:05 PM

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

**Starmix** · June 12th 2009, 10:34 PM

Merci.....!

)

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

**Soroban** · June 13th 2009, 03:38 AM

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.

**Soroban** · June 13th 2009, 04:10 AM

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.

**mr fantastic** · June 13th 2009, 04:23 AM

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.

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