a rectangular garden is to be fenced along four sides and in the middle it is to be divided into 2 equal areas.if 300meters of fencing materials is available, find the maximum possible area that can be fenced.

make a sketch, so you will have 2 longs sides and 3 short sides parallel with the divider.

Let the long sides by y, and the short sides be x
3x + 2y = 300
y = (300 - 3x)/2 = 150 - 1.5x
Area = xy
= x(150-1.5x)
= 150x - 1.5x^2
= -1.5(x^2 - 100x + 2500 - 2500)
= -1.5(x-50)^2 + 3750

Max area = 3750 m^2

check: x = 50, y = 75
area = 3750, 3x+2y= 150+150 = 300

To find the maximum possible area that can be fenced, we need to consider the constraints of the problem.

Let's assume the length of the rectangle is L and the width is W.

Given that the garden is to be fenced along all four sides and divided into two equal areas, we can set up the following equation for the total amount of fencing material used:

2L + 2W = 300

To divide the garden into two equal areas, we divide it along the length with a fence. This means the width of each divided section is still W, but the length of each section is L/2.

Now, let's substitute the value of L/2 for the length of one section in terms of W, and substitute it in the equation for the total amount of fencing material:

2(L/2) + 2W = 300
L + 2W = 300

Now, we need to isolate L in terms of W:

L = 300 - 2W

To find the maximum area, we need to maximize the product of the length and width.

Area(A) = L * W = (300 - 2W) * W

To find the maximum area, we can use calculus by differentiating the area equation with respect to W and finding the critical points.

dA/dW = 300 - 4W

Setting dA/dW to zero:

300 - 4W = 0
4W = 300
W = 75

So, when W = 75, the area is maximized. Substituting this value back into the equation for L:

L = 300 - 2(75) = 150

Therefore, the maximum possible area that can be fenced is:

Area = L * W = 150 * 75 = 11,250 square meters.