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December 3rd, 2015, 04:03 AM   #1
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Help to find the dimensions

Hello All:
I have this Question. I really don't know how to answer it. Can someone help me and thanks for all.

The Question is:

Poster is to be designed with 50 in2(Square) of printed type, 4 inch margins on both the top and the bottom, and 2 inch margins on each side. Find the dimensions of the poster which minimize the amount of paper used.

Last edited by skipjack; December 3rd, 2015 at 05:22 AM.
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December 3rd, 2015, 04:20 AM   #2
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For a printed area $x$ wide by $y$ high, we require a printed area $xy = 50$.

The paper width is $x+4$ (the printed width plus a border of 2 at each side).
The paper height is $y+8$ (the printed height plus a border of 4 each at the top and bottom).

The total area of paper used is then $f(x)=(x+4)(y+8) = xy + 8x + 4y + 32$, but we have $xy=50$ so $y = {50 \over x}$. Putting both of these into the equation gives
$$f(x) = 50 + 8x + {200 \over x} + 32 = 82 + 8x + {200 \over x}$$

The minimum value of this function specifies the printed width we require. From that, you can work out the paper size.
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Last edited by skipjack; December 3rd, 2015 at 05:11 AM.
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December 3rd, 2015, 04:32 AM   #3
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Quote:
Originally Posted by v8archie View Post
For a printed area $x$ wide by $y$ high, we require a printed area $xy = 50$.

The paper width is $x+4$ (the printed width plus a border of 2 at each side).
The paper height is $y+8$ (the printed height plus a border of 4 each at the top and bottom).

The total area of paper used is then $f(x)=(x+4)(y+8) = xy + 8x + 4y + 32$, but we have $xy=50$ so $y = {50 \over x}$. Putting both of these into the equation gives
$$f(x) = 50 + 8x + {200 \over x} + 32 = 82 + 8x + {200 \over x}$$

The minimum value of this function specifies the printed width we require. From that, you can work out the paper size.
Thanks for your help. But there is many things that are not shown clear in your text (the words and some symbols like this \$) right ?

Last edited by skipjack; December 3rd, 2015 at 05:13 AM.
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December 3rd, 2015, 05:19 AM   #4
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Is it readable now? The minimum (for $x$ > 0) occurs for x = 5.
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December 3rd, 2015, 06:37 AM   #5
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Quote:
Originally Posted by skipjack View Post
Is it readable now? The minimum (for $x$ > 0) occurs for x = 5.
Thank you a lot .
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December 3rd, 2015, 06:47 AM   #6
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Quote:
Originally Posted by v8archie View Post
For a printed area $x$ wide by $y$ high, we require a printed area $xy = 50$.

The paper width is $x+4$ (the printed width plus a border of 2 at each side).
The paper height is $y+8$ (the printed height plus a border of 4 each at the top and bottom).

The total area of paper used is then $f(x)=(x+4)(y+8) = xy + 8x + 4y + 32$, but we have $xy=50$ so $y = {50 \over x}$. Putting both of these into the equation gives
$$f(x) = 50 + 8x + {200 \over x} + 32 = 82 + 8x + {200 \over x}$$

The minimum value of this function specifies the printed width we require. From that, you can work out the paper size.
So It can't solve ? Because I don't have the width?

Last edited by skipjack; December 3rd, 2015 at 09:29 AM.
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December 3rd, 2015, 08:25 AM   #7
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Yes, you can solve for $x$. Once you have $x$, you can calculate $y$ and anything else you want.
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December 4th, 2015, 07:33 AM   #8
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Quote:
Originally Posted by v8archie View Post
Yes, you can solve for $x$. Once you have $x$, you can calculate $y$ and anything else you want.
Does in finally x = -2 ?? or = 2 ?

Last edited by ahmed2009; December 4th, 2015 at 07:37 AM.
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December 4th, 2015, 07:53 AM   #9
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Skipjack got $x=5$. That's probably correct.
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December 5th, 2015, 01:25 AM   #10
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Quote:
Originally Posted by v8archie View Post
Skipjack got $x=5$. That's probably correct.
But he didn't show how he solve it ...
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