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 December 3rd, 2015, 04:03 AM #1 Member   Joined: Nov 2015 From: saudi arabia Posts: 30 Thanks: 5 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.
 December 3rd, 2015, 04:20 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra 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 from ahmed2009 Last edited by skipjack; December 3rd, 2015 at 05:11 AM.
December 3rd, 2015, 04:32 AM   #3
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Joined: Nov 2015
From: saudi arabia

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Quote:
 Originally Posted by v8archie 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.  December 3rd, 2015, 05:19 AM #4 Global Moderator Joined: Dec 2006 Posts: 21,020 Thanks: 2255 Is it readable now? The minimum (for$x$> 0) occurs for x = 5. Thanks from ahmed2009 December 3rd, 2015, 06:37 AM #5 Member Joined: Nov 2015 From: saudi arabia Posts: 30 Thanks: 5 Quote:  Originally Posted by skipjack Is it readable now? The minimum (for$x$> 0) occurs for x = 5. Thank you a lot . December 3rd, 2015, 06:47 AM #6 Member Joined: Nov 2015 From: saudi arabia Posts: 30 Thanks: 5 Quote:  Originally Posted by v8archie 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.  December 3rd, 2015, 08:25 AM #7 Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra Yes, you can solve for$x$. Once you have$x$, you can calculate$y$and anything else you want. Thanks from ahmed2009 December 4th, 2015, 07:33 AM #8 Member Joined: Nov 2015 From: saudi arabia Posts: 30 Thanks: 5 Quote:  Originally Posted by v8archie 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.  December 4th, 2015, 07:53 AM #9 Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra Skipjack got$x=5$. That's probably correct. Thanks from ahmed2009 December 5th, 2015, 01:25 AM #10 Member Joined: Nov 2015 From: saudi arabia Posts: 30 Thanks: 5 Quote:  Originally Posted by v8archie Skipjack got$x=5\$. That's probably correct.
But he didn't show how he solve it ...

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