A home owner wants to fense 3 sides of his rectangular yard. She has 120 feet of fencing available. What dimensions will yield the maximum area?

Question

A home owner wants to fense 3 sides of his rectangular yard. The fourth side will be the side of the house. She has 120 feet of fencing available. What dimensions will yield the maximum area?

Answer 1

Let x = length of side parallel to the house

     y = length of each side perpendicular to the house

 

Then, x + 2y = 120.  So, y = 60 - (1/2)x

 

Area = xy = x[60 - (1/2)x]

 

The graph of the area function is a parabola opening downward, so there is a maximum.  By symmetry, the x coordinate of the maximum point occurs halfway between the x-intercepts.  Since the x-intercepts are 0 and 120, the maximum occurs when x = 60.

 

So, to maximize the area, the side parallel to the house should be 60 ft long and the other two sides should each have length

30 ft.  The maximum area is then (60)(30) = 1800 ft².

 

  

Answer 2

The rectangle that has the greatest area for a specified perimeter is the square. 

120 / 3 = 40

 

The dimensions should be 40 x 40 ft producing and area of 1600 ft²