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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

Q: 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?

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Polygon

It is called a polyhedron.

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You divide the area of the shaded region by the area of the full circle. For example, if the radius of the shaded region is 2 meters, the probability would be 4pi / 36pi, or 1/9. If the shaded region is a 'slice' of the circle, the chance is just the fraction of the circle which the 'slice' is.

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true lalu

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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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a veterinarian uses 600 feet of chain-link fence to enclose a rectangular region. It is subdivided evenly into two smaller rectangles by placing fence parallel to one of the sides. what is the width (w) as a function of the length (L)? (w=?) what is the total area as a function of (L)? (A=?) what are the dimensions that produce the greatest enclosed area? EXPLENATION PLEASE

Your question does not contain enough information to be answered. Region 2 where?

Head region