A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. Three hundred feet of fencing is used dimensions of the playground that maximize the total enclosed area. What is the maximum area?

The smaller dimension is

feet

Step-by-step explanation:

Let x represent the length of the side not parallel to the partition. Then the length of the side parallel to the partition is ...

  y = (300 -2x)/3

And the enclosed area is ...

  A = xy = x(300 -2x)/3 = (2/3)(x)(150 -x)

This is the equation of a parabola with x-intercepts at x=0 and x=150. The line of symmetry, hence the vertex, is located halfway between these values, at x=75.

The maximum area is enclosed when the dimensions are ...

  50 ft by 75 ft

That maximum area is 3750 square feet.

Comment on the solution

The generic solution to problems of this sort is that half the fence (cost) is used in each of the orthogonal directions. Here, half the fence is 150 ft, so the long side measures 150'/2 = 75', and the short side measures 150'/3 = 50'. This remains true regardless of the number of partitions, and regardless if part or all of one side is missing (e.g. bounded by a barn or river).

deez nu tz

step-by-step explanation: