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mandilynn22
01.06.2021 •
Mathematics
You work at a farm that wants to build three identical rectangular corrals in a row against the side of a barn using 112 ft of fencing. No fence is needed next to the barn itself. What dimensions (for each individual corral) will maximize the corral areas
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Ответ:
x = 14 ft
y = 56 ft
A(max corral) = 784/3 = 261.33 ft²
Step-by-step explanation:
Let´s call y the side parallel to the barn, then only one y is going to be fenced.
If we are going to divide the area in three identical corrals we need 4 times x ( the other side perpendicular to the barn)
The perimeter of the rectangular area ( divide in three identical corrals)
112 = y + 4*x or y = 112 - 4*x
A (r) = x*y
Area as a function of x is
A(x) = x* ( 112 - 4*x) A(x) = 112*x - 4*x²
Tacking derivatives on both sides of the equation
A´(x) = 112 - 8*x A´(x) = 0 112 - 8*x = 0
x = 112/8
x = 14 ft
And y = 112 - 4*x y = 112 - 56
y = 56 ft
A(max) = 14 * 56 = 784 ft²
A(max corral) = 784/3 = 261.33 ft²
How do we know the area is maximum, tacking the second derivative
A´´(x) = - 8 A´´(x) < 0
Then the function A(x) has a maximum at the point x = 14
Ответ:
for eg:- 2×3=6
this means we are adding 2 3 times
ie, 2+2+2=6