A farmer is building a fence to enclose a rectangular area consisting of two separate regions. The four walls and one additional vertical segment (to separate the regions) are made up of fencing, as shown below. If the farmer has 234 feet of fencing, what are the dimensions of the region which enclose the maximal area?

9514 1404 393

  58.5 ft by 39 ft

Step-by-step explanation:

Let x represent the length of the two horizontal segments. Then the three vertical segments will be ...

  (234 -2x)/3

The total enclosed area is the product of these dimensions:

  A = (x)(234 -2x)/3

  A = (2/3)(x)(117 -x)

This is the equation of a downward-opening parabola with zeros at x=0 and x=117. The maximum of the parabola will be on the line of symmetry, halfway between these zeros. The value of x there is ...

  x = (0 +117)/2 = 58.5

The lengths of the vertical segments are ...

  (2/3)(117 -58.5) = 2/3(58.5) = 39

The dimensions of the region enclosing the maximum area are 58.5 ft by 39 ft. The additional vertical segment is 39 ft.

Comment on maximum area problems

You may have noticed that the total length of the fence allocated to the long sides (2×58.5 = 117) is half the total length of fence and is equal to the total length of fence allocated to the short sides (3×39 = 117).

This relationship is true in all rectangular fencing problems where the area is being maximized for a given total fence length. It doesn't matter how many partitions there are in either direction: the total of horizontal lengths is equal to the total of vertical lengths.

it is x=-4

step-by-step explanation: