Posted by thomas on .
Suppose you have 168 meters of fencing with which to make two sidebyside rectangular enclosures against an existing wall. If the rectangular enclosures are adjacent and of the same depth, what is the maximum are that can be enclosed?

advanced math 
Damon,
total length = x
depth = y
fencing needed = x + 3 y = 168
A = x y
A = (1683y)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
(y28)^2 = (1/3)(A2352)
so vertex is at A = 2352 square meters
by the way
x = 168 3y = 84 
advanced math 
tchrwill,
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. 
advanced math 
tchrwill,
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.