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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

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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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