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
You have 32 feet of fencing and want to determine the maximum area that can be enclosed with the fencing to make a flower garden. You do not want to deal with fractions or decimals, so you are limiting the dimensions to whole numbers. Also, the area is to be rectangular (or square) in shape.
2 answers
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.
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.
1 answer