Posted by thomas on Thursday, November 3, 2011 at 6:44pm.
total length = x
depth = y
fencing needed = x + 3 y = 168
A = x y
A = (168-3y)y = 168 y - 3 y^2
3 y^2 - 168 y = -A
y^2 - 56 y = -A/3
y^2 - 56 y + 784 = -A/3 +784
(y-28)^2 = -(1/3)(A-2352)
so vertex is at A = 2352 square meters
by the way
x = 168 -3y = 84
Let the sides parallel to the given boundary be x and the sides perpendicular to the boundary be y.
Then, 2x + 3y = 168 and A = 2xy.
With y = (168 - 2x)/3
A = (336x - 4x^2)/3
dA/dx = 112 - 8x/3 = 0
336 = 8x making x = 84 and y = 28
The area A = 2352m^2.
Alternatively:
Considering all rectangles with a given perimeter, one side being provided by a straight given boundry, which one encloses the largest area?
Letting P equal the given perimeter and "x" the short side of the rectangle, we can write for the area A = x(P - 2x) = Px - 2x^2.
Taking the first derivitive and setting equal to zero, dA/dx = P - 4x = 0, x becomes P/4.
With x = P/4, we end up with a rectangle with side ratio of 2:1.
.....The short side is P/4.The traditional calculus approach would be as follows.
.....The long side is (P - 2(P/4)) = P/2.
Therefore, it can be unequivicably stated that of all possible rectangles with a given perimeter, one side being a given external boundry, the rectangle with side ratio of 2:1 encloses the maximum area.