A pig farmer wants to enclose a rectangular area and then divide it into six pens with fencing parallel to one side of the rectangle (see the figure below). There are 610 feet of fencing available to complete the job. What is the largest possible total area of the six pens?
699
Let's assume the length of the rectangular area is L and the width is W.
To enclose the rectangular area, we need to use fencing for 2 lengths and 2 widths, which gives us a total of 2L + 2W feet of fencing.
Additionally, we need to use three additional pieces of fencing (parallel to the width) to divide the area into six pens. Since there are six pens, each with two sides, we need a total of 6 * 2 = 12 feet of fencing for this purpose.
Therefore, the total amount of fencing used is 2L + 2W + 12.
Given that the farmer has 610 feet of fencing available, we can write the equation:
2L + 2W + 12 = 610.
Now, let's solve this equation for one variable in terms of the other and find the maximum area.
2L + 2W = 610 - 12,
2L + 2W = 598,
L + W = 299.
To solve for the maximum area, we can use the fact that A = L * W.
Let's express one variable in terms of the other and substitute it into the area equation:
W = 299 - L,
A = L * (299 - L).
To find the maximum area, we need to find the vertex of the quadratic equation A = L * (299 - L).
The vertex of a quadratic equation in the form A = -L^2 + 299L is given by the formula L = -b/(2a).
In our case, a = -1 and b = 299. Substituting these values into the formula, we get:
L = -299 / (2 * -1),
L = 299 / 2,
L = 149.5.
Since L represents the length, which cannot be a decimal, the closest whole number we can use is 150.
Now, substituting the value of L back into the equation W = 299 - L, we get:
W = 299 - 150,
W = 149.
Therefore, the dimensions of the rectangular enclosure that will maximize the area are L = 150 feet and W = 149 feet.
Finally, we can calculate the maximum total area by multiplying the length of each pen by the width of the enclosure:
Total Area = 6 * (L * W),
Total Area = 6 * (150 * 149),
Total Area = 6 * 22,350,
Total Area = 134,100 square feet.
Thus, the largest possible total area of the six pens is 134,100 square feet.
To find the largest possible total area of the six pens, we can use the quadratic optimization technique.
Let's assume the dimensions of the rectangular area are length (L) and width (W).
We know that the perimeter of a rectangle is given by the formula:
P = 2L + 2W
In this case, the perimeter is given as 610 feet:
610 = 2L + 2W
We also know that the rectangular area is divided into six equal pens, meaning there will be five dividing fences parallel to one side of the rectangle.
Let's calculate the length of each dividing fence:
Length of one dividing fence (F) = L/6
Now, let's express the width of the rectangular area in terms of L and F.
The width is divided into six equal segments, and since there are five dividing fences, the total width (W) can be expressed as:
W = 6F
Substituting W = 6F into the perimeter equation:
610 = 2L + 2(6F)
610 = 2L + 12F
Now we have two equations:
1. 610 = 2L + 12F
2. W = 6F
To solve for F, we can rearrange equation 1 to solve for L in terms of F:
L = (610 - 12F) / 2
L = 305 - 6F
Now we substitute this L value into equation 2:
W = 305 - 6F
To find the largest possible total area of the six pens, we need to maximize the area, which is given by:
Area = L * W
Substituting the expressions for L and W:
Area = (305 - 6F) * (305 - 6F)
Expanding this equation:
Area = 305^2 - 12F * 305 + 12F * 305 - 36F^2
Area = 305^2 - 36F^2
Now, we have the area of the six pens as a quadratic equation. To find the maximum area, we can graph the equation or use calculus techniques.
Using calculus, we differentiate the Area equation with respect to F:
d(Area)/dF = -72F
Setting this derivative equal to zero to find the critical point:
-72F = 0
F = 0
Now, we need to ensure that this is a maximum point. To determine this, we can take the second derivative:
d^2(Area)/dF^2 = -72
Since the second derivative is negative (-72), this confirms that F = 0 is a maximum point.
Now we substitute F = 0 back into the original equation for Area:
Area = 305^2 - 36(0)^2
Area = 305^2
Therefore, the largest possible total area of the six pens is 305^2 square feet.