A farmer wishes to put a fence around a rectangular field and then divide the field into three rectangular plots by placing two fences parallel to one of the sides. If the farmer can afford only 1600 yards of fencing, what dimensions will give the maximum rectangular area? yd(smaller value)? yd (larger value)?

If the field has dimensions x and y, and two extra length of y are used inside the field,

2x+4y = 1600
so, x = 800-2y

the field's area is

a = xy = (800-2y)(y)
= 2y(400-y)

This is a parabola with vertex at y=200, so the maximum area is achieved when the field is 400x200

Well, let's put on our imaginary farmer hat and get cracking on this problem! To maximize the area, we need to find the dimensions that will give us the largest possible rectangular area within our fence.

So, let's say the length of the field is x yards. We'll also have two fences parallel to x, so the width of each rectangular plot will be (x/2).

Now, we need to take into account the fencing requirements. The perimeter of our fence will be x + 2(x/2) + 2(x/2), which is equal to 1600 yards.

Simplifying, we get:

x + x + x = 1600
3x = 1600
x = 533.33 (rounded to two decimal places)

Now, we have the length of the field, which is x = 533.33 yards. And we know that the width of each rectangular plot will be (x/2), which is (533.33/2) = 266.67 yards.

So, the dimensions that will give us the maximum rectangular area are approximately 266.67 yards (smaller value) by 533.33 yards (larger value).

Now go forth, my farmer friend, and embrace those optimal dimensions!

To find the dimensions that will give the maximum rectangular area, we can use the concept of optimization. Let's denote the length of the rectangular field as L and the width as W.

Given that the farmer can afford only 1600 yards of fencing, the total length of the five fence segments would be:

2L + 3W = 1600

To maximize the rectangular area, we need to express the area, A, in terms of a single variable. We know that the area of a rectangle is given by the formula:

A = L * W

Since we have an equation for the total length of the fences in terms of L and W, we can isolate L in terms of W and substitute it into the equation for the area. Let's solve for L:

2L + 3W = 1600
2L = 1600 - 3W
L = (1600 - 3W) / 2

Now, substitute this expression for L into the area formula:

A = L * W
A = ((1600 - 3W) / 2) * W
A = (1600W - 3W^2) / 2

To find the maximum area, we can find the critical points by taking the derivative of the area function with respect to W and setting it to zero. Let's find dA/dW:

dA/dW = (1600 - 6W) / 2
dA/dW = 800 - 3W

Setting this equal to zero, we get:

800 - 3W = 0
3W = 800
W = 800 / 3 ≈ 266.67

Now, substitute this value of W back into the equation for L:

L = (1600 - 3W) / 2
L = (1600 - 3(800 / 3)) / 2
L = (1600 - 800) / 2
L = 800 / 2 = 400

So, the dimensions that give the maximum rectangular area are approximately:
Width (smaller value): 266.67 yards
Length (larger value): 400 yards

To find the dimensions that will give the maximum rectangular area, we can use the concept of optimization.

Let's assume the length of the field is x yards and the width is y yards. We know that the perimeter of the field is given by:

2x + 3y = 1600 (since there are two lengths and three widths)

We need to express the area of the rectangular field in terms of a single variable. The area, A, of a rectangle is given by:

A = x*y

We can solve the equation for x by writing it in terms of y:

2x = 1600 - 3y
x = (1600 - 3y)/2

Substituting this expression for x into the equation for the area, we have:

A = ((1600 - 3y)/2)*y
A = (800y - 3y^2)/2

To find the maximum area, we need to find the value of y that maximizes this expression.

One way to do this is by taking the derivative of A with respect to y and setting it equal to zero:

dA/dy = 800 - 6y = 0
6y = 800
y = 800/6
y = 133.33 (approximately)

Since we are dealing with lengths, we need to choose the closest whole number value for y. So, we can choose y = 133 yards.

Now, we can substitute this value of y back into the equation for x:

x = (1600 - 3(133))/2
x = (1600 - 399)/2
x = 601/2
x = 300.5 (approximately)

Again, we need to choose the closest whole number value for x, so x = 300 yards.

Therefore, the dimensions that will give the maximum rectangular area are approximately 300 yards by 133 yards.