A farmer wants to fence an area of 6 million square feet in a rectangular field and then divide it in half with a fence parallel
1 answer:
If the rectangular field has notional sides X and y then it has area
A(x) =xy { =6•10^6 sq ft }
The length of fencing required, if
x
is the letter that was arbitrarily assigned to the side to which the dividing fence runs parallel, is:l (x) = 3x +2y
It matters not that the farmer wishes to divide the area into 2 exact smaller areas.
Assuming the cost of the fencing is proportional to the length of fencing required, then
C(x)=a L (x)
To optimise cost, using the Lagrange Multiplier
λ
, with the area constraint :
So the farmer minimises the cost by fencing-off in the ratio 2:3, either-way
You might be interested in
Answer:
7. cos(X) = 3/5 = 0.6
cos(Y) = 4/5 = 0.8
8. cos(X) = 15/17 ≈ 0.8824
cos(Y) = 8/17 ≈ 0.4706
9. cos(X) = 1/2 = 0.5
cos(Y) = (√3)/2 ≈ 0.8660
Step-by-step explanation:
The trigonometric ratio for cosine is given as follows;
7. The adjacent leg length to ∠X = 27
The length of the hypotenuse side of the triangle = 45
∴ cos(X) = 27/45 = 3/5 = 0.6
Cos(X) = 3/5 = 0.6
The adjacent leg length to ∠Y = 36
∴ cos(Y) = 36/45 = 4/5 = 0.8
cos(Y) = 4/5 = 0.8
8. The adjacent leg length to ∠X = 15
The length of the hypotenuse side of the triangle = 17
∴ cos(X) = 15/17 ≈ 0.8824
The adjacent leg length to ∠Y = 8
∴ cos(Y) = 8/17 ≈ 0.4706
9. The adjacent leg length to ∠X = 13
The length of the hypotenuse side of the triangle = 26
∴ cos(X) = 13/26 = 1/2 = 0.5
Cos(X) = 1/2 = 0.5
The adjacent leg length to ∠Y = 13·√3
∴ cos(Y) = 13·√3/26 = (√3)/2 ≈ 0.8660
cos(Y) = (√3)/2 ≈ 0.8660.
Answer:
c) 120 ft
Step-by-step explanation:
Let's consider the rhombus has 4 sides, A, B, C, and D.
To find the length of each side, let's first find the length AE.
From the diagram, AE is half of AC and AC = 30 ft.
Therefore,
AE = ½ * 30
AE = 15 ft
Let's find the length AD, since we are looking for the distance around the perimeter.
We are told the rhombus is formed by four identical triangles.
Therefore the distance around the perimeter would be: AD+AD+AD+AD=
30 ft + 30 ft + 30 ft + 30 ft
= 120 ft
The distance around the perimeter of the garden is 120 ft