October 24th, 2012, 06:58 PM  #1 
Senior Member Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0  Rectangles
A rectangle is bounded by the XAxis on the bottom and bounded by the YAxis on the left side. The upper right corner of the rectangle is bounded by the graph of y= (6x)/(2) = 1/2x + 3 What lengh, X and width, Y, should the rectangle have so that its area is a maximum? y = 1/2x +3 A = xy. Find max area using 1st derV test 
October 24th, 2012, 09:24 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: Rectangles
You have: hence: Note: conceptually, this is the same as the Farmer and his fence...and may be solved without calculus, although I suggest you do use the calculus, since this is what you are studying, and it is simpler too, at least computationally. I would write: Now do the differentiation thing... 
October 25th, 2012, 05:39 PM  #3 
Senior Member Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0  Re: Rectangles
and I am getting 3X is this right? Me and Calculus are not enjoying each others company lately 
October 25th, 2012, 06:37 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: Rectangles
Yes, that's right. Now, what does the second derivative tell you about the extremum at this critical value?

October 25th, 2012, 07:03 PM  #5 
Senior Member Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0  Re: Rectangles
ok 2nd derV would be 3x set to zero so 3x = 0 x =3 if i did table it would be on the open intervals of (inf, +3)U(+3,+inf) thus we choose our test values I will simply choose 0 and +10 thus we get f"(0) > 0 so here we are at concave upward and f"(+10) < 0 so here we are at concave down. so we know that there is an inflection at x = +3 and the actual point would be plug +3 into original equation and get the inflection point of (3,11/2) did I make mistakes here Mark? please review this carefully and let me know if I did a mishap/misstep thanks 
October 25th, 2012, 07:07 PM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: Rectangles
The second derivative is the constant 1, and since is it less than zero, we know the parabola is concave down for all x, and so the extremum we got from the first derivative must be a maximum. Sometimes is is easier to use the second derivative test for relative extrema, than to use the first derivative test.

October 25th, 2012, 09:09 PM  #7 
Senior Member Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0  Re: Rectangles
oh can u show me where I went wrong? could u set up the first derivative and the second derivative for me so I can see? thank you 
October 25th, 2012, 09:27 PM  #8 
Senior Member Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0  Re: Rectangles
I am still a little confused on what the max area is. I went to do another problem and got mixed up when I came back here what is my max area? x= y= ? 
October 25th, 2012, 09:49 PM  #9  
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: Rectangles Quote:
and so: Quote:
 
October 25th, 2012, 10:25 PM  #10 
Senior Member Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0  Re: Rectangles
wow i must have just ff'd somewhere and got off the path..lol has that ever happened to you? got off the path? anyway, I also made a type o because my sheet says that y = 1/2 x + 3 (not +1/2) therefore ... y = 1/2(+3 + 3) ... y = 1.5 + 3 ... y = +3/2 and then I shall calculate again using the Amax but with the correct +3/2 

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