Math Math???????????????????????Need Help?

Question

A rectangular field will be fenced on all four sides. There will also be a line of fence across the field, parallel to the shorter side. If 900 m of fencing are available, what dimensions of the maximum area randy can enclose?

Answer 1

Find the total length of fencing. You fence the perimeter which is:
2W + 2L

You fence the inside, which is parallel to the short side, or width:
W

Total fencing is 2W + 2L + W = 900 m

3W + 2L = 900
L = (900 - 3W) / 2

Area = L * W
But Area = [(900 - 3W) / 2] * W from above.
A = 450W - (3/2)W^2

Now you can maximize area by taking the derivative of Area, and setting the derivative equal to zero. Solve for W and you have the width you are looking for:

A' = 450 - 3W = 0
3W = 450
W = 150 meters

Solve for Length by plugging in width above:

3W + 2L = 900 -----> 3(150) + 2L = 900
2L = 450
L = 225 meters

Answer 2

Intuition told me that it should be a square 180m x 180m (32,400 sq. m), but the fence across the middle affects things.

There are 3 sections of short fence (3n). And there are two sections of long fence (900-3n)/2.

Then the area is given by product of these two lengths, namely:
n * (900 - 3n)/2.

Or this can be rewritten as 450n - 3n²/2.

To find the maximum value, take the derivative and set it to zero...

So you have 450 - 3/2*2n, or

450 - 3n = 0
450 = 3n
150 = n

So the short length is 150m and the long length is 225m.

Interestingly, 450m is devoted to the 3 short fences (3x150) and 450m is devoted to the 2 long fences (2x225).

The dimensions of the maximum enclosed area are 150m x 225m (a total of 33,725 sq. m. which is bigger than my initial guess).

Answer 3

Hoping you already know that the maximum area of a "rectangle" is when that rectangle is a square then you have 5 equal sides of fencing which means you divide 900 by 5 and the dimension are 180m by 180m.

Answer 4

If length = L and width = W, then

3W + 2L = 900

If the field is square, then W = L and

5L = 900
L = 180

If the field is not square, then maybe there is not enough information given to solve the problem?

Answer 5

you have 900 = 3x + 2y...
area = xy, y = 1.5x - 450.
area = 1.5x^2 + 450x.
take it's derivative, and you get:
3x = 450,
x = 150, so y = 225.

you've got 225*150 = 33750 m squared.so, 150, 150, 150, 225, 225. meaning the rectangle is 150*225.

Answer 6

2x+3y= 900

x*y must be maximized.

So x= 1.5(300-y)

450y-1.5y^2=900

1.5y^2 -450 y+900=0.

The first answer is correct (and the second). And mine : )