# If x ft. of stonewall is used as one side of the field, express the area enclosed as a function of x according to the problem below.A farmer has 600ft of woven wire fencing available to enclose a...

If x ft. of stonewall is used as one side of the field, express the area enclosed as a function of x according to the problem below.

A farmer has 600ft of woven wire fencing available to enclose a rectangular field and to divide it into three parts by two fences parallel to one end.

If x ft. of stonewall is used as one side of the field, express the area enclosed as a function of x when the dividing fences are parallel to the stone wall.

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we have 600 ft of wire to be used as the fence and separation.

If ove side is x Then we need 3x because one of the sides is stone wall

let the other side be y , then we need 2y

==> 3x + 2y = 600

==> 2y = 600 - 3x

==> y= 300 - (3/2)x ........(1)

Now we need the area of the rectangular field:

==> A = x*y

==> A = x*[300-(3/2)x]

=300x - (3/2)x^2

**==> The are A = 300x - 3x/2 **

Since the field is rectangular, we'll establish the dimensions length and width as x and y.

Now, we know, from enunciation that there are available 600 ft wire to enclose the field and to build more wire walls, namely 2 inner walls, parallel to the stone wall.

So, the total amount of 600 ft could be expressed as:

3x + 2y = 600 (1)

We did not put 4x because one wall is made of stone and we did not put 2x because we have 2 more inner wire walls, besides the end wall.

To calculate y with respect to x, we'll subtract 3x both sides, in (1).

2y = 600 - 3x

We'll divide by 2:

y = 300 - 3x/2 (2)

Now, we'll express the area enclosed:

A = length*width

A = x*y

We'll substitute y by (2):

A = x*(300 - 3x/2)

We'll remove the brackets and we'll have:

**A = -3x^2/2 + 300x **

Let ABCD be the rectangle.

Let the stone wall be along AB and let AB = x.

Then CD , being the opposite side of the rectangleAB, has the length x.

Let KLand MN be two fences parallel fences to AB. K and and M are on BC and L and N are on DA.

Now The fence of 6000 is utised for BC,CD,AD , KL and MN.

Also AB= CD= KL = MN , as KL and MN are || to AB (and So CD also.)

AB side is not faenced, as there is a stone wall.

Therefore AD = BC = (600- (CD+KL+MN) )/2= (600-3x)/2.

So the area of the rectangle ABCD = AB * BC = x*(600-3x)/2..

Therefore the area function, A(x) = x(600-3x)/2 sq units.