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November 27, 2014

November 27, 2014

Posted by **Anonymous** on Thursday, November 3, 2011 at 11:02am.

- geometry -
**Reiny**, Thursday, November 3, 2011 at 12:37pmnot 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

- geometry -
**tchrwill**, Friday, November 4, 2011 at 10:52amConsidering 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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