Answer:
Shorter side = 755 yd
longer side = 1510 yd
A(max) = 1140050 yd²
Step-by-step explanation:
The owner has 3020 yd of fencing
Lets assume:
x the shorter side of the rectangle ( we will use fencing in two sides of length x )
y the longer one
A area of the rectangle and P the perimeter of the rectangle ( we have only three sides covered by fencing material)
We have: A = x * y P = 2 * x + y ⇒ y = P - 2*x ⇒ y = 3020 - 2* x
A (x) = x * ( 3020 - 2*x) ⇒ A(x) = 3020 * x - 2* x²
Taken derivative
A´(x) = 3020 - 4 * x
If A´(x) = 0 3020 -4*x = 0 ⇒4*x = 3020 x = 755 yd
If we take second derivative A´´(x) = -4
so A´´(x) < 0 so there is a maximun in point x = 755
Then
Rectangle dimensions :
x = 755 yd ⇒ y = 3020 - 2 * x ⇒ y = 3020 - 2 * (755) y = 1510 yd
Maximum area is : A(max) = 1510 * 755 ⇒ A(max) = 1140050 yd²
The largest area he can enclose is 1,013,374 yards², across 4 equal sides of 1,006.66 yards.
Since the owner of the Rancho Grande has 3020 yd of fencing with which to enclose a rectangular piece of grazing land situated along the straight portion of a river, to determine, if fencing is not required along the river, what are the dimensions of the largest area he can enclose, the following calculations must be performed:
Therefore, the largest area he can enclose is 1,013,374 yards², across 4 equal sides of 1,006.66 yards.
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