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Ron will use all 900 meters of fence and wants to enclose a rectangular region that contain four rectangular pens within it what will be the area for each pen?
My drawing shows this equation. (A = XY--remember this! )
5X + 2Y = 900
solve for Y
Y = ( 900 - 2X)/2
substitute back in; remember X*Y
5X(900 - 2X)/2
4500X - 10X^2/2
2250X - 5X^2
d/dx( 2250X - 5X^2)
2250 - 10X
2250 = 10X
X = 225
as you see 900/4 is 225, so this is the optimal area for each pen
What else can I help you with?
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Optimization is all about finding equations that involve your variables, and manipulating those equations to meet the stated constraints. The first step is to find two or more equations involving one or more of your variables. Here, we have cost and two separate lengths. Let's call the long side of each of the small congruent rectangles "l" and the short side "w." You know that you will have four of these rectangles, and because of the situation and the units you know that you will be measuring perimeter. Therefore, the sum of the perimeters of all four congruent rectangles is 4(2l+2w) or 8l+8w. I'm unclear from your question whether Ron will be using the 900 meters to form the four rectangles, or to form the rectangles and to surround them with fencing. Can you give me more information?
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