Posted by **thomas** on Thursday, November 3, 2011 at 6:44pm.

Suppose you have 168 meters of fencing with which to make two side-by-side 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**, Thursday, November 3, 2011 at 6:55pm
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

- advanced math -
**tchrwill**, Friday, November 4, 2011 at 10:46am
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**, Friday, November 4, 2011 at 10:49am
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.

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