Samples of size n= 240 are randomly selected from the population of numbers (O through 20)
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The dimensions of the rectangular pen should be 15 by 20 feet and the maximum area is 1200 square feet.
Let the area be y .
Area = (base) × (height)
Base = 2x
Height = h
Let the area of the rectangular pens be y .
∴ y = 2xh
Perimeter of all the fencing = 4x+3h
∴ 4x+3h = 120
now we solve for h
3h = 120-4x
h = 40 - 4/3 x
Now we will substitute this value in the above first equation:
y = 2xh
or, y = 2x (40 - 4/3 x)
or, y = 80x - 8/3 x²
Now for the maximum area we have to find the first order differentiation of y
now,
dy /dx = 80 - 16/3 x
At dy/dx = 0 we get the value of x for which y is maximum.
80 - 16/3 x = 0
or, - 16/3 x = -80
or, x = 15 feet
Hence height = 40 - 4/3 x = 40 - 20 = 20feet
Maximum area = 2xh = 2×15×40 = 1200 square feet
The dimensions of the rectangular pen should be 15 by 20 feet and the maximum area is 1200 square feet.
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Answer:
Fn= 174.9 N : Magnitude of the net force the people exert on the donkey.
Step-by-step explanation:
We find the components of the forces in x-y-z
Force of Jack in z =F₁z=90.5 N in direction (+z)
Force of Jill in x = F₂x= -82.3*cos45°= - 58.19 N (-x)
Force of Jill in y =F₂y=-82.3*sin45°= + 58.19 N (+y)
Force of Jane in x =F₃x=125*cos45°= + 88.4 N (+x)
Force of Jane in y =F₃y=125*sin45°= + 88.4 N (+y)
Calculating of the components of the net force the people exert on the donkey.
Fnx= F₂x+F₃x=( - 58.19+ 88.4 )N=30.2N (+x)
Fny= F₂y+F₃y=( 58.19+88.4 ) = 146.59 N (+y)
Fnz =F₁z=90.5 N (+z)
Calculating of the magnitude of the net force the people exert on the donkey.