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April 23, 2014

April 23, 2014

Posted by **joe** on Wednesday, February 2, 2011 at 8:49pm.

- math -
**Ms. Sue**, Wednesday, February 2, 2011 at 9:18pmThe largest area would be a square, measuring 20 feet on each side.

- math -
**joe**, Friday, February 4, 2011 at 3:22pmcan you explain how you got the answer?

- math -
**joe**, Friday, February 4, 2011 at 3:22pmcan you explain how you got the answer?

- math -
**tchrwill**, Sunday, February 13, 2011 at 9:30amA popular problem in recreational mathematics seeks the dimensions of a rectangular garden that maximizes the garden area for a given perimeter of fencing. It is very often posed in two different versions. One asks for the dimensions of a rectangular garden with maximum enclosed area for a fixed perimeter while the other asks for the same thing while using the wall of a barn as one side of the garden. Not too surprisingly, the answers are different.

Lets first address the garden with fencing on all four sides.

What are the dimensions of a rectangular garden with a fixed perimeter or 60 ft. that maximizes the enclosed area?

Considering all rectangles with the same perimeter, the square encloses the greatest area.

Proof: Consider a square of dimensions x by x, the area of which is x^2. Adjusting the dimensions by adding

"a" to one side and subtracting "a" from the other side results in an area of (x + a)(x - a) = x^2 - a^2. Thus, however small the dimension "a" is, the area of the modified rectangle is always less than the square of area x^2.

Therefore, the dimensions of the garden with maximum area for the given perimeter, p, of fencing is p/4 by p/4.

Given a length of fencing of 60 feet, the garden dimensions become 15 by 15 ft. for an area of 225 sq.ft.

Lets now assume that we have a linear barn wall to act as one edge of the rectangle we wish to enclose.

Letting the length of the side perpendicular to the barn wall equal "x", we can express the dimensions of the garden area by A = x(p - 2x) = px - 2x^2. Taking the first derivitive of A and setting it equal to zero, we derive p - 4x = 0 or 4x = p making x = p/4. While this appears to be the same answer as for the 4 sided garden, with x = p/4, the other dimension becomes p - 2(p/4) = p - p/2 = p/2. While the area of the garden fenced on all four sides was A = p^2/16, the area of the garden with fencing on only three sides becomes (p/4)(p/2) = p^2/8, literally twice the area of the square garden fenced on all four sides. Thus, for our simple example presented earlier, with a fence length of 60 ft., the maximum enclosed area using the barn wall as one side becomes 60^2/8 = 450 sq.ft.

(Without the use of the calculus, you can arbitrarily select values of x and compute A and plot the results on graph paper. The resulting plot will show the area being a maximum when x = p/4.)

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