Sat Dec 04, 2010 6:10 pm by tartle |
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write 271 as the sum of positive real numbers so as to maximize their product. |
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Mon Jan 03, 2011 1:13 pm by s.b. |
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2+3+4+5+6+7+8....
note:0 is not applied cuz then product becomes 0. 1 does not change the product...am i right? |
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Sun May 29, 2011 7:48 pm by bds021 |
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135.5+135.5 |
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Sat Jun 04, 2011 7:03 am by Unni |
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3+2+2+2+2+2.......
Product = 3 x 134th power of 2 |
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Sat Jun 04, 2011 7:05 am by Unni |
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3+2+2+2+2+2.......
Product = 3 x 134th power of 2 |
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Sun Jun 19, 2011 6:05 pm by DiamondSoul |
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It's 2.71 repeated 100 times. |
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Wed Jul 06, 2011 4:28 pm by cat |
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[quote="DiamondSoul"]It's 2.71 repeated 100 times.[/quote]
This appears to be correct, but needs a little explanation.
([i]n[/i]+1)*([i]n[/i]-1) = [i]n[/i]^2-1, which is less than [i]n[/i]^2. This shows that uniform values adding to a given sum make the largest product. Therefore, using [i]a[/i] for 271, the product [i]y[/i] of [i]x[/i] uniform values can be written as:
[i]y[/i] = ([i]a[/i]/[i]x[/i])^[i]x[/i] = e^[[i]x[/i]*(ln[i]a[/i]-ln[i]x[/i])]
The derivative is:
[i]dy[/i]/[i]dx[/i] = [([i]a[/i]/[i]x[/i])^[i]x[/i]]*[ln[i]a[/i]-(1+ln[i]x[/i])] = [([i]a[/i]/[i]x[/i])^[i]x[/i]]*[ln([i]a[/i]/[i]x[/i])-1]
At the maximum product, the derivative equals zero, so that:
ln([i]a[/i]/[i]x[/i]) = 1;
[i]a[/i]/[i]x[/i] = e;
[i]x[/i] = [i]a[/i]/e = 271/2.718... = 100 to the nearest integer. |
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