Posted by Anonymous on Thursday, November 3, 2011 at 11:02am.
not much of a problem here ...
the largest rectangle is obtained when that rectangle is a square, so
4 equal sides equal 32 feet,
each side is 8 feet
area is 64 square ft
Considering all rectangles with a given perimeter, which one encloses the largest area?
The traditional calculus approach would be as follows.
Letting P equal the given perimeter and "x" the short side of the rectangle, we can write for the area A = x(P - 2x)/2 = Px/2 - x^2.
Taking the first derivitive and setting equal to zero, dA/dx = P/2 - 2x = 0, x becomes P/4.
With x = P/4, all four sides are equal making the rectangle a square.
.....The short side is P/4.
.....The long side is (P - 2(P/4))/2 = P/4.
Therefore, it can be unequivicaly stated that of all possible rectangles with a given perimeter, the square encloses the maximum area.
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