A rectangular field is to be subdivided in 6 equal fields. There is 1200 feet of fencing available.
Find the dimensions of the field that maximizes the total area. (List the longer side first)
Width = feet
Length = feet
What is the maximum area ? Area =
how are the 6 fields to be arranged?
2x3:
9x+8y = 1200
a = 6xy = 6x(1200-9x)/8
max area at x = 200/3
1x6:
12x+7y = 1200
a = 6xy = 6x(1200-12x)/7
max area at x = 50
Note that the fencing is divided equally among lengths and widths.
To find the dimensions of the field that maximize the total area, we can use calculus to solve this optimization problem.
Let's assume the width of each small field is w, and the length is L.
Since we have 6 equal fields, the total length of the fencing required is 6 times the width plus 4 times the length. So we have the equation:
6w + 4L = 1200
To find the dimensions that maximize the area, we need to express the area in terms of a single variable.
The total area of the field is given by:
A = 6wL
Since we have an equation that relates w and L (6w + 4L = 1200), we can solve for one variable in terms of the other and substitute it in the area equation.
Rearranging the fencing equation, we get:
6w = 1200 - 4L
Dividing both sides by 6, we obtain:
w = (1200 - 4L) / 6
Now, we can substitute this value of w in the area equation:
A = 6[(1200 - 4L) / 6] L
Simplifying this expression gives:
A = (1200 - 4L)L
To find the maximum area, we need to find the critical points of this equation. Since this equation is quadratic, we can find the maximum by taking the derivative and setting it equal to zero.
dA/dL = 1200 - 8L
Setting this derivative equal to zero, we can solve for L:
1200 - 8L = 0
8L = 1200
L = 150
Now that we have the value of L, we can substitute it back into the equation for w:
w = (1200 - 4 * 150) / 6
w = (1200 - 600) / 6
w = 100
So, the dimensions that maximize the total area are width = 100 feet and length = 150 feet.
To find the maximum area, we substitute these values into the area equation:
A = (1200 - 4 * 150) * 150
A = 600 * 150
A = 90,000 square feet
Therefore, the maximum area is 90,000 square feet.