Question

A rancher needs to enclose two adjacent rectangularâ corrals, one for cattle and one for sheep. if the river forms one side of the corrals and 180 yd of fencing isâ available, find the largest total area that can be enclosed.

Answer 1

Let B = breadth makes the three sides of 2 corrals &
Let L = Length parallel to the river,

Here, L + 3B = 180 yd
So, L = 180 - 3B

We know that the area of the rectangle,
A = L × B
   =  (180 - 3B) × B
   = 180 B - 3$B^{2}$
 
We know that quadratic equation has symmetry,
So, For maximum area Breadth B = $\frac{-2a}{b}$

Here, a = -3 & b 180

So, B = $\frac{-180}{2(-3)}$
          = 30 yd

So, L = 180 - 3B
         = 180 - 3(30)
         = 180 - 90
         = 90

So, Maximum Area = L × B
                               = 90 × 30
                               = 2700 square yards

Therefore, the correct answer is 2700 square yards.

Answer 2

Refer to the diagram shown below.

x =  the width of each rectangular area.
y = the length of each rectangular area.

The amount of fencing available is 180 yards, therefore
4x + 2y = 180
or
2x + y = 90
y = 90 - 2x            (1)

The total enclosed area is
A = 2xy
   = 2x(90 - 2x)
 A = 180x - 4x²      (2)

In order to maximize A, 
A'(x) = 180 - 8x = 0
x = 180/8 = 22.5 yd
y = 90 - 2x = 45.0 yd

To verify that A will be maximum, A''(x) should be negative.
A'' = -8  (verified).

The maximum area is
Amax =  2*22.5*45.0 = 2025.0 yd²

Answer: 2025 yd²