A rectangular rice land is to be fenced along 4 sides and in the middle to divide it into to 2 equal areas. If 200m of fencing is available, what is the maximum area to be fenced?

Assume that length l is greater than or equal to width w. Assert that the dividing fencing must be equal to the width (ie it must be perpendicular) by reasoning that if it were at any other angle, it would be longer, and the length would be reduced.

A = w * l
200 = 3w + 2l
.: A = w*(200-3w)/2
or A = 100 w - (3/2) w^2

Take the derivative and find when this is 0.
0 = (d/dw) (100 w-(3/2)w^2)

Solve for w, use to determine A

where did you get the 3w

The 200m fencing is used for two length-sides (l), two width-sides (w) and the middle divider (w).

This: 200 = 2l + 3w

To find the maximum area to be fenced, we need to determine the dimensions that will allow us to maximize the area while using the given amount of fencing.

Let's assume that the width of the rectangular rice land is "w" and the length is "l". Since we have to divide the land into two equal areas, the length of one portion will be (l/2) and the other portion will also have a length of (l/2).

The fencing required for the four sides can be calculated as follows:
- The top and bottom sides: 2w
- The left and right sides: l + (l/2) + (l/2) = l + l = 2l

Since we have 200m of fencing available, we can write the equation:
2w + 2l = 200

Now, let's solve this equation for either "w" or "l". For simplicity, let's solve it for "w":
2w = 200 - 2l
w = (200 - 2l)/2
w = 100 - l

To find the maximum area, we need to find the dimensions that will maximize the product of length and width (lw).

Substituting the value of "w" from the above equation, we have:
Area (A) = l * (100 - l)

Now, let's take the derivative of A with respect to l to find the critical points:
dA/dl = 100 - 2l

Setting dA/dl = 0, we have:
100 - 2l = 0
2l = 100
l = 50

To ensure that this is a maximum point, we can take the second derivative:
d^2A/dl^2 = -2

Since the second derivative is negative, it confirms that this is a maximum point. Hence, when l = 50, we can find the maximum area by substituting the value of l back into the area equation:
A = l * (100 - l)
A = 50 * (100 - 50)
A = 50 * 50
A = 2500 square meters

Therefore, the maximum area to be fenced is 2500 square meters.