A farmer wants to fence a small rectangular yard next to a barn. Fence for side parallel to the barn will cost 55 per foot and the fence for the other two sides will cost 20 per foot. The farmer has a total of 1950 dollars to spend on the project. Find the dimensions for the yard that will have the largest possible area.

Question

Axle

Mathematics

8 August, 02:23

A farmer wants to fence a small rectangular yard next to a barn. Fence for side parallel to the barn will cost 55 per foot and the fence for the other two sides will cost 20 per foot. The farmer has a total of 1950 dollars to spend on the project. Find the dimensions for the yard that will have the largest possible area.

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Answers (2)

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Lily Yoder · Answer 1

We conclude that a dimension of the yard

17·25=425

Step-by-step explanation:

We know that the farmer has a total of 1950 dollars to spend on the project. Also we know that fence for side parallel to the barn will cost 55 per foot and the fence for the other two sides will cost 20 per foot.

We conclude that the farmer must fence one side parallel to the barn will cost 55 per foot, and the fence for the other two sides will cost 20 per foot.

We get

55+2·20=55+40=95

The farmer has a total of 1950 dollars, we have

1950:95=20.52

As the other two sides are cheaper to build, then they will be longer than the side parallel to a barn.

We conclude that the dimensions for the yard that will have the largest area

25 foot whic cost 20 per foot

17 foot whic cost 55 per foot

We get:

17·55+2· (25·20) = 935+1000=1935 < 1950

We conclude that a dimension of the yard

17·25=425

Natalie Hodge · Answer 2

The dimension should be 17.73ft by 24.37ft

Step-by-step explanation:

The area of a rectangle is given as;

A = L*B ... 1

The fence would have three sides

L = a length parallel to the barn

2B = the two other sides

Total cost = 1950

Total cost of fencing the yard;

55L + 20 (2B) = 1950

Making B the subject of formula

B = (1950-55L) / 40 ... 2

Substituting eqn 2 into eqn 1

A = L * (1950-55L) / 40

A = (1950L - 55L^2) / 40 ... 3

Therefore, at maximum area, dA/dL = 0

Differentiating equation 3;

dA/dL = (1950-110L) / 40 = 0

1950 - 110L = 0

L = 1950/110

L = 17.73ft

Substituting the value of L into equation 2

B = (1950-55 (17.73)) / 40

B = 24.37ft

Therefore, the length parallel to the barn should be 17.73ft and the other two sides should be 24.47ft each.

Confirmation;

Area = 17.73 * 24.37 = 432.1ft^2

Cost = 17.73*55 + 24.37 (2*20) = 1949.95 < / = 1950 dollars.

Confirmed