Afarmer wishes to erect a fence enclosing a rectangular area adjacent to a barn which 24 feet long. his plan for the fenced area indicates the fencing in bold black lines; it shows also that the barn will be used as part of the fencing on one side. find the largest are, , that can be enclosed if 80 feet of fencing material is available.

a) =674 sq. ft.

b) =672 sq. ft.

c)= 676 sq. ft.

d)=670 sq. ft.

  c)  676 sq. ft.

Step-by-step explanation:

With the 24 ft barn, the total perimeter can be 24+80 = 104 feet. This allows the farmer to make a square pen 104/4 = 26 feet on a side, for an area of ...

  (26 ft)² = 676 ft²

Comment on "fence" problems

These fence problems have different solutions for different ratios of "barn" length to "fence" length. When the barn length is more than 1/2 the length of the fence, the optimum shape is a rectangle in which half the fence is used parallel to the barn. (The full length of the barn is not used.)

When the barn length is between 1/2 and 1/3 the fence length, the optimum shape is a rectangle with one side equal to the barn length.

When the barn length is less than 1/3 the fence length, as here, the optimum shape is a square, with the barn side being augmented by fence as required. Here, 2 ft of fence is added to the 24 ft barn to make a square 26 ft on each side.

step-by-step explanation:

given perimeter: 28cm

length: 9cm

atp,

2(l+w)=p of the reactangle

2(9+x)=28

18+2x=28

2x=28-18

x=10/2

x=5

therefore width is 5 cm