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October 24th, 2012, 05:58 PM   #1
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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
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October 24th, 2012, 08:24 PM   #2
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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...
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October 25th, 2012, 04:39 PM   #3
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Re: Rectangles

and I am getting 3-X


is this right?

Me and Calculus are not enjoying each others company lately
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October 25th, 2012, 05:37 PM   #4
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Re: Rectangles

Yes, that's right. Now, what does the second derivative tell you about the extremum at this critical value?
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October 25th, 2012, 06:03 PM   #5
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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
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October 25th, 2012, 06:07 PM   #6
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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.
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October 25th, 2012, 08:09 PM   #7
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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
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October 25th, 2012, 08:27 PM   #8
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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=
?
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October 25th, 2012, 08:49 PM   #9
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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:

and so:



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 , we then have:





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October 25th, 2012, 09:25 PM   #10
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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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