Homework Help: Calculus

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

A man has 340 yards of fencing for enclosing two separate fields, one of which is to be a rectangle twice as long as it is wide, and the other, a square. The square field must contain at least 100 square yards and the rectangular one must contain at least 800 yards.

(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 12:18pm
the 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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