Consider the following problem: a farmer with 750 feet of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens

14,062.5ft^2

Step-by-step explanation:

A farmer has a given length of fencing. Since this fencing is used to enclose a rectangular area, before it is then divided it into four pens using the same provided fencing material. This will definitely lead to five lengths of fencing which come along one side of the rectangle to form four pens.

Let us assume that the length of the each of the five sides be xft.

Then let the length of each of the two sides in the be yft.

Since the total length of fencing available is 750 feet.

Then we can have an equation of such:

5x+2y=750

Making y the subject

2y = 750-5x

y = 750-5x/2. 1

The area of the rectangular region is given by:

A = xy. 2

Combining (1) in (2) to find the area

A(x)= x. (750−5x)/2

=750x−5x^2/2 (*)

Going by the means of first diffential calculus.

A(x)=750x−5x^2/2

A'(x) = 750-10x/2

Setting A'(x) = 0

750-10x/2 = 0

750-10x = 2×0

750-10x = 0

750= 10x

x = 750/10

x = 75

Now that we have x = 75

We can now substitute into equation (*)

A(x) = 750×75-5×75^2/2

= (56250-28125)/2

= 28126/2

= 14062.5

Hence the rectangular region of maximum area is

14062.5ft^2

ab will be equal to