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 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 X-Axis on the bottom and bounded by the Y-Axis on the left side. The upper right corner of the rectangle is bounded by the graph of y= (6-x)/(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: $A=xy$ $y=\frac{6-x}{2}$ hence: $A(x)=\frac{x(6-x)}{2}$ 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: $A(x)=\frac{1}{2}$$6x-x^2$$$ 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 3-X 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 3-x set to zero so 3-x = 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:
 Originally Posted by mathkid 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
You correctly found:

$f'(x)=3-x$ and so:

$f''(x)=-1$

Quote:
 Originally Posted by mathkid 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= ?
Having determined a maximum occurs at $x=3$, we then have:

$x=3$

$y=\frac{6-3}{2}=\frac{3}{2}$

$A_{\text{max}}=3\cdot\frac{3}{2}=\frac{9}{2}$

 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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