A farmer decides to enclose a rectangular garden, using the side of a barn as one side of the rectangle. what is the maximum area that the farmer can enclose with 40ft of fence? What should th dimensions of the garden be in order to yield this area?
For this problem I struggled on how to figure in the side of the barn into my formula, which would have been:
2w+2l=40
A=l x w
A=(40-w)w
and then I would have turned it into a quadratic formula and solved.
Thx for any help on showing me how to figure in the side of the barn into my formula.
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A 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.)
Before we leave this interesting exercise, it is worth pointing out that the question is usually posed by specifying that the garden area be rectangular. Once in a while, the question will ask you to define the garden shape that will maximize the garden area with the given length of fencing. While not immediately obvious, an octagon will enclose more area for the fixed perimeter than a square. Carrying this to its extreme, the circle will enclose the maximum area for a given perimeter. The dimensions of the square are p/4 by p/4. The radius of a circle of perimeter p is r = p/2Pi. This makes the area of the square p^2/16 and the area of the circle p^2/4Pi = p^2/12.56. Clearly, the circle enlcloses ~27% more area for the same perimeter, ~286 sq.ft. compared to 225 sq.ft. for the square.
Of course, the next logical question you might now ask is what shape will enclose the maximum area for the given perimeter using the barn wall as one edge of the garden. Quite logically, we can be reasonably safe in concluding that the answer is some portion of a circle. This one is more easily computed by assuming percentages of the circular perimeter of a full circle and computing the area of the circle minus the area of the segment inside the barn wall. For instance, if I assume that the perimeter of a circle of radius R is 3/4 outside the barn wall, I can equate the perimeter p to 2PiR(240/360) = 60 making the radius of this partial circle 14.32 ft. The net area of the circular area outside the barn wall then becomes A = PiR^2 - R^2(µ - sinµ)/2 where R^2(µ - sinµ)/2 is the area of the segment of the circle within the barn wall and µ is the central angle subtended by the circle chord coincident with the barn wall. For the 3/4 circle we assumed, a central angle of 90º, the circle radius becomes 14.324 ft. and the area becomes 518.581 sq.ft. The perimeter of the 3/4 circle outside the barn wall is our 60 ft. Computing other radii and areas for larger central angles ultimately leads us to the surprising conclusion that the shape of the garden that maximizes the garden area for the given perimeter of fencing is a semi-circle with the diameter coincident with the barn wall.
