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posted by  A mom on 11/2/2009 1:24:32 PM  |  status: Closed  |  Earned Karma: 10

Please show me the correct way to work this in DETAIL. Lifesaver awarded!

Course Textbook Chapter Problem Needs by
Calculus N/A N/A N/A 11/2/2009 at 10:00:00 PM
Question Details:
Find the area of the largest rectangle with one corner at the origin, the opposite corner in the first quadrant on the graph of the line f(x) = 32-4x, and sides parallel to the axes.
The area equals:____________?

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Oracle
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posted by anonymousxyz on 11/2/2009 2:02:28 PM  |  status: Live
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Response Details:





The area of the rectangle is: 



     =  xy 

     =  x( 32  -  4x ) 

     =  32x  -  4x2


Maximize A:


A'  =  32  - 8x


A'  =  0  =>

32  -  8x  =  0  =>

8x  =  32  =>

x  =  32 / 8  =>

x  =  4  units =>


y  =  32  -  4( 4 )  =>

y  =  32  -  16  =>

y  =  16 units


The area of the largest rectangle is:


A  =  ( 4 )( 16 )  =  64 units2


Apprentice
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posted by gammabeta on 11/2/2009 2:16:05 PM  |  status: Live
Asker's Rating: Lifesaver   
Response Details:
Here is the graph:
Area Function:
Let A'(x)=0, so
32 - 8x = 0
   x=4 is a critical number.  So we test this critical number with the endpoints of our closed interval.
By the Extreme Value Theorem, the absolute maximum area is 64 units2.
Tags: Optimization
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