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September 1, 2014

September 1, 2014

Posted by **Amy** on Sunday, February 5, 2012 at 7:30pm.

(a.) If X is the width of the rectangular field, what are the maximum and minimum values of X?

(b.) What is the greatest number of square yards that can be enclosed in the two fields?

If you know how to answer even part of this question, that would be so helpful. Thank you so much :x :X

- Calculus -
**Anonymous**, Monday, February 6, 2012 at 12:18pmthe square has to be at least 10x10, using 40 yds of fence. That leaves 240 yds for the rectangle, so X+2X<=120, or X <= 40.

If X = 40, 2X = 80, so area = 3200

If 2X^2 = 800, X=20, so

20 <= X <= 40

As X increases from 20 to 40, the area increases from 800 to 3200

total area of square + rectangle can be as much as 3300

If the rectangle is 20x40, that uses up 120 yds of fence, leaving 220 for the square, or 55x55, with area 3025. Total area with maximum square is thus 3825.

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