In this section, we will explore what happens to the area of a rectangle when the perimeter of the rectangle is a fixed value.
Recall
Recall that the perimeter is the distance around all four sides of the rectangle.
In the Explore This, you might have noticed:
A rectangle has a perimeter of \(12\). In the table of values, the width of the rectangle starts with \(1\) and increases by \(1\). Calculate the length, area, and the first differences in area to complete the table. How do the first differences suggest when the maximum area of the rectangle occurs?
Since the width increases by \(1\), the first differences are the differences between consecutive area values.
The table suggests the maximum area is \(9\) when the width and the length are both \(3\).
Before the maximum, the area is increasing. The first differences are positive.
After the maximum, the area is decreasing. The first differences are negative.
A rectangle has a perimeter of \(14\). Complete the table of values then graph the relation of the width with the area. How does the graph tell you when the maximum area of the rectangle occurs?
The greatest area that occurs in the table of values is \(12\). It occurs twice.
Notice the symmetry of the graph. This hints that the maximum area might be halfway between the two highest points.
The width and length would both be \(3.5\), and the area would be \(12.25\). This would be the maximum area of the rectangle.
For a rectangle enclosed on four sides with a fixed perimeter, the maximum area occurs when the length and the width of the rectangle are equal.
A rectangular yard is to be enclosed on four sides using \(60\) m of fencing. What would be the dimensions of the yard with the greatest possible area?
Since the fencing for the yard is on all four sides of the rectangle, the greatest possible area will occur when the width and the length are equal or the yard is a square.
Let \(s\) represent the side length of the square yard.
Since the perimeter is \(60\) m, we can write this as an equation and solve for \(s\):
Therefore, the dimensions of the yard with the greatest possible area are \(15\) m by \(15\) m.