Question

More optimization problems!

A rectangular page is to contain 100 square inches of print. The margins on each side are 1 inch. Find the dimensions of the page such that the least amount of paper is used.

Answer 1

start with some equations

Answer 2

Okay, well, I know that the area is going to be:
A=x*y ; and I think the page looks like this:

Answer 3

Am, I kind of right? :x

Answer 4

These problems are kinda confusing.

Answer 5

if the dimensions of the page are x, y. then the text will have area (x-2)(y-2)=100

Answer 6

Ahh, the -2 is just because 2 inches are taken off each side for margins, got it.

Answer 7

So, I think next I would have to solve for one of the variables?

Answer 8

yess...

Answer 9

A=x*y
Constraint --> A=x*y=100;
So y = 100-x ? :D

Answer 10

the area of the print was
(x-2)*(y-2)=100
[solve this for y]


the area of the page was
A=x*y

Answer 11

Hmm...
xy-2x-2y+4 = 100; I have no idea how my mind is becoming flustered by such a simple thing, ugh. =_=

Answer 12

@MoonlitFate don't foil it. We can solve for y much easier:
\[\Large (x-2)*(y-2)=100\]
what happens if we divide both sides by (x-2)

Answer 13

Wow, I fail. Lol.\[y-2 = \frac{ 100 }{ x-2 }\]

Answer 14

+2 no -2 on the right.

Answer 15

Correct, now add 2 to both sides (or is that what you meant with your last post..?)

Answer 16

\[y = \frac{ 100 }{ x+2 }+2\]

Answer 17

how did that two in the denominator change sign?

Answer 18

Apparently my brain is turning to mush.
\[y = \frac{ 100 }{ x-2 }+2\]
Am I good now?

Answer 19

that is good,

now plug this into

A=x*y

Answer 20

All right.\[A=x(\frac{ 100 }{ x-2 }+2)\]\[=\frac{ 100x }{ x-2 }+2x\]

Answer 21

ok, good,

can you simplify that fraction a little?

Answer 22

\[-50+2x?\]

Answer 23

um , you cant do that/

Answer 24

maybe just leave the fraction as it was before,

can you find the derivative of A

Answer 25

Oh, oops. D: Sorry. Yeah, I can find the derivative.

Answer 26

\[A'(x) = 2-\frac{ 200 }{ (x-2)^2 }\]

Answer 27

good

Answer 28

Now, since we want to minimize A, you need to find where A'(x) = 0, so solve for x
\[\Large A'(x) = 2-\frac{ 200 }{ (x-2)^2 } = 0\]
We might have to check the sign of the second derivative to confirm it's a minimum, but for now, just solve for x.

Answer 29

All right.

x=12 or x = -8

Answer 30

Great stuff,
now only one of those values for x is going to the length of a rectangle, right?

Answer 31

plug this back into
(x-2)*(y-2)=100
and solve for y

and your done .

Answer 32

Well, you can't have a negative value for height. :p

Answer 33

thats right

Answer 34

I got that y would also be equal to 12. :)

Answer 35

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