Suppose you want to fence in a field,
and then subdivide it into three equal parts.
You have 2400 yards of fencing material to
use. What should be the dimensions of one
of the three smaller fields, if you want to
fence in the maximum area?
I will assume that the three smaller parts are equal rectangles.
Let the width of one of the smaller rectangles is x
and its length is y
Make a sketch and you should see that
our original field is 3x by y
and perimeter is 6x + 4y
so 6x + 4y = 2400
3x + 2y = 1200
y = (1200 - 3x)/2 = 600 - 3x/2
area = 3xy
= 3x(600 - 3x/2)
= 1800x - (9/2)x^2
d(area)/dx = 1800 - 9x
= 0 for a max of area
9x = 1800
x = 200 and y = 600 - 3(200)/2 = 300
So each of the smaller fields is 200 by 300 yards
Note that the maximum area is achieved when the fencing is divided equally among the lengths and widths: 1200 each.
To determine the dimensions of one of the three smaller fields that would maximize the area, we can use optimization techniques by applying calculus.
Let's denote the length of the outer fence as 'L' and the width as 'W'. There are three smaller fields, so each smaller field has the same dimensions, and let's denote them as 'l' and 'w'.
We know that the total amount of fencing material available is 2400 yards. Thus, the perimeter of the outer fence is given by:
P = 3L + W
Now, since we want to maximize the area, we need to express it in terms of a single variable. In this case, the area of one of the smaller fields can be calculated as the product of its length and width:
A = l * w
To eliminate one variable, we can express one variable in terms of the other using the given information. We know that the length of the outer fence is made up of three lengths of one small field and two widths, so we have:
L = 3l + 2w
Now, we can rewrite the equation for the perimeter using the expression for L:
P = 3L + W
P = 3(3l + 2w) + W
P = 9l + 6w + W
We also have the equation for the perimeter in terms of the length and width of one small field:
P = 2l + 2w
Since the total amount of fencing material is fixed at 2400 yards, we can equate the equations for P:
9l + 6w + W = 2l + 2w
Next, we need to solve the equation to find the relationship between l and w. By rewriting the equation, it becomes:
7l + 4w = -W
We can express l in terms of w as:
l = (-4w - W) / 7
Now, we substitute this equation for l into the expression for the area A:
A = l * w
A = [(-4w - W) / 7] * w
We have expressed the area as a function in terms of w. To maximize A, we can take the derivative of A with respect to w, set it to zero, and solve for w. This will give us the critical points, one of which will correspond to the maximum area.
After finding the critical point(s), substitute the value(s) of w into the equation for l to find the corresponding length.
Once you have the values of l and w, you can calculate the dimensions of one of the three smaller fields.