A developer wants to envlose a rectangular grassy lot that borders a city street for parking.

Question

college algebra.

 

A developer wants to enclose a rectangular grassy lot that borders a city street for parking. If the developer has 336 feet of fencing and does not fence the side along the street, what is the largest area that can be enclosed and what are the dimensions that give this maximum area?

Answer 1

Hello, Courtney

The largest area that can be enclosed is (84)(168)=14,112

The dimensions are length= 168, width= 84

This is how:

perimeter/4= width 

perimeter/2= length

 

This works when you need to fence only three sides, though.  Try it out with smaller numbers.  Let say you have 20 feet, and you need to fence three sides.  Remember, opposite sides of a rectangle are congruent.  These are the possibilities

lenghth           width (must be double)          Area.      Perimeter

   10.                 5.                                       50.            20

     8.                 6.                                        48.            20

      6.                7.                                         42.           20

      4.                 8.                                        32.            20 

      2.                 9.                                        18.             20

     12.                4.                                         48.             20

     14                  3.                                        42.             20

      16                2.                                         32.             20

       18.              1.                                          18.            20

 

Sorry about the crooked numbers, I am using an ipad.  As you can see Courtney, the one that gives us the greatest area is the one the has a length that is 1/2 the perimeter and a width that is 1/4 the perimeter .  In this case, the perimeter was 20, so our ultimate length should be 20/2=10, and our ultimate width should be 20/4=5.  

 

Ok, good problem.. It was fun!! Please rate my answer:)

 

D. Y. Taylor

Answer 2

P=2l+2w

336=2l+2w

168=l+w

l=168-w

A=lw

A=(168-w)(w)

A=168w-w²

this is a parabola whose vertex is (h,k)

h=(-b)/(2a)

h=-168/(2)(-1)

h=84

h is the maximizing width and k is the area

(168)(84)-84²=14,112-7056=7056

k=7056 square feet

The largest area that can be enclosed is always enclosed  by a square.

If you were enclosing all four sides, you would have an 84 by 84 square which would maximize the area.

However, one side is not going to be fenced so take the 84 feet you would have used on this side and add it to the opposite side.

You have a rectangle that is 84 feet by 168 feet by 84 feet by "the street".

A=84*168

A=14,112 square feet

Notice that this rectangle is made up of two squares, each 84 by 84 !!

84*84=7056 and 2*7056=14,112