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anonymous
 5 years ago
a page of print is to contain24in^2 of printed region,a margin of 1and1/2 at the bottom,and a margin of 1in.at the sides..what are the dimensions of the smallest page that will fill these requirements?
anonymous
 5 years ago
a page of print is to contain24in^2 of printed region,a margin of 1and1/2 at the bottom,and a margin of 1in.at the sides..what are the dimensions of the smallest page that will fill these requirements?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://tutorial.math.lamar.edu/Classes/CalcI/Optimization.aspx See example 6

radar
 5 years ago
Best ResponseYou've already chosen the best response.0A great reference. Thanks godnod. Let me see if I can put it together and held ronadelossantos. Let y equal one dimension of the printed material. Let x = the other dimension of the printed material. It is stated that the printed material is 24 sq. inches. Thus: xy=24. We will use that information later. One dimension of the paper (including the margins) will be x+1 and the other dimension is y+2. Returning to the xy=24 we can say y=24/x. We now have enough information to solve.

radar
 5 years ago
Best ResponseYou've already chosen the best response.0\[A=(x+1)((x/24)+2)\] This becomes: \[24x/x+2x+24/x+2\] \[24+2x+24/x+2\] \[2x+24/x+26=A\] \[2x+24x ^{1}+26=Area\] Differntiating and set zero for min/max

radar
 5 years ago
Best ResponseYou've already chosen the best response.0A' = 0\[A'=224x ^{2}=0\]

radar
 5 years ago
Best ResponseYou've already chosen the best response.0\[24=2x ^{2}\] \[x ^{2}=12\] \[x=2\sqrt{3}\] one dimension solve for y y=24/x \[y=24/(2\sqrt{3}=12/\sqrt{3}=4\sqrt{3}\]
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