David has 520 yards of fencing to enclose a rectangular area

David has 520 yards of fencing to enclose a rectangular area. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area? A rectangle that maximize the enclosed area has a length has a length of yards and a width of yards.

Solution

Given that David has 520 yards of fencing to enclose a rectangular area.

It means circumference of rectangular area is 520 yards.

Let length of the rectangle is X yards

width of the rectangle is Y yards.

Circumference of rectangular area is 520 yards

2(X+Y)=520

X+Y=260

The sum of the length and width of a rectangle is 260.

The X and Y values should be taken in which it has a maximum area.

XY= MAXIMUM.

By making some trials, it is found that X=131 yards, Y=129 Yards.

Which satisfies X+Y=260 yards and XY has the maximum rectangular area.

Area= XY=(131)(129)

=16,899 yd²

Therefore, Length of a rectangle= 131 yards

Width of a rectangle= 129 yards

Maximum Area of that rectangle= 16,899 yd²