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 February 17th, 2014, 08:39 PM #1 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 find max(a+b+c) $a\in Z$ $b,c \in R$ $a>b , \,\, \text{and} \,\, a>c$ given : $a+2b+3c=6 -----(1)$ $abc=5-----(2)$ if: $k=\min(a)$ find: $\max(a+b+c)$
 February 25th, 2014, 03:38 AM #2 Senior Member   Joined: Jan 2007 From: India Posts: 161 Thanks: 0 Re: find max(a+b+c) Just curious to know on the answer of this question..can someone help please. Cheers, Arun
 February 26th, 2014, 02:31 PM #3 Global Moderator   Joined: Dec 2006 Posts: 21,026 Thanks: 2257 It's an easy question. You're looking for (a, b, c) = (10, -1/2, -1).
February 26th, 2014, 06:48 PM   #4
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Quote:
 Originally Posted by skipjack It's an easy question. You're looking for (a, b, c) = (10, -1/2, -1).
Thanks skipjack... could you please let me know how you arrive at the solution?

 February 26th, 2014, 07:28 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: find max(a+b+c) I guess I'm missing something but it seems max(a + b + c) can be made arbitrarily large by choosing a arbitrarily large. a = 11, b = (-55 - ?1705))/44, c = -20/(55 + ?1705) works and the sum a + b + c is just over 8.6. I'm not sure what "if: k = min(a)" is about.
 February 27th, 2014, 01:06 AM #6 Global Moderator   Joined: Dec 2006 Posts: 21,026 Thanks: 2257 Obviously, "if: k = min(a)" isn't what was really meant. I assumed it was intended that a has its minimum value (which is 10), in which case, a + b + c has a maximum value (as only two values are possible).
March 1st, 2014, 04:53 AM   #7
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Re: find max(a+b+c)

Quote:
 Originally Posted by greg1313 I guess I'm missing something but it seems max(a + b + c) can be made arbitrarily large by choosing a arbitrarily large. a = 11, b = (-55 - ?1705))/44, c = -20/(55 + ?1705) works and the sum a + b + c is just over 8.6. I'm not sure what "if: k = min(a)" is about.
Greg, is there any method you used to arrive at the value of a=11.. is 8.6 the max of (a+b+c)..if you could give the steps, would be helpful. Thanks.

 March 1st, 2014, 09:21 AM #8 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 Re: find max(a+b+c) Arun, why don't you send Abert Teng a PM...
 March 1st, 2014, 08:16 PM #9 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: find max(a+b+c) I chose a = 11. It's then possible to find values for b and c that fulfill the two given equations.
 March 3rd, 2014, 11:59 PM #10 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 Re: find max(a+b+c) we know : $a\in N$ $c=\dfrac {6-2b-k}{3}=\dfrac {5}{kb}---(1)$ if : $b \in R$ then : $(k^2-6k)^2\geq120k$ or: $k(k-6)^2\geq 120$ $\therefore\, \min(a)=k=10$ put to(1)  we get : $b=\dfrac {-1}{2} or,\, \dfrac {-3}{2}$ so : $(a,b,c)=(10,\dfrac {-1}{2}, -1)$ or : $(a,b,c)=(10,\dfrac {-3}{2}, \dfrac {-1}{3})$ $\therefore\, \max(a+b+c)=10-\dfrac {1}{2}-1=\dfrac{17}{2}$

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