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 January 30th, 2013, 09:24 PM #1 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 find minimum x,y>0 and 9x+4y=2005 $\text find :min(\frac{1}{x}+\frac {1}{y})\text$ Ans : $\frac {5}{401}$
January 31st, 2013, 07:13 AM   #2
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Re: find minimum

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 January 31st, 2013, 07:42 AM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: find minimum 335/26866 if x and y are integers; x=133 and y=202
 February 2nd, 2013, 11:07 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2203 For the exact answer, (x, y) = (401/3, 401/2).
 February 3rd, 2013, 04:51 AM #5 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 Re: find minimum $\text \frac {1}{x}+\frac{1}{y}=\frac{2005+5y}{2005y-4y^2}=g(y)\text$ find the derivatives of g(y)=g'(y) let g'(y)=0 we get : $\text 20y^2+8\times2005y-2005^2=0\text$ (10y-2005)(2y+2005)=0 $\text \therefore y=\frac {401}{2}\,\,,\,\, x= \frac {401}{3}\,\, (here\,\, x,y >0)\text=$ $\text \,\, and min(\frac {1}{x}+\frac{1}{y})= =\frac {5}{401}\text$

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