Sometimes there are natural borders that can be used to help enclose an area, such as a wall or a body of water.

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In cases like this, we can use the savings to enclose a larger area.

In Explore This 2, you might have noticed:

• The sum of the \(3\) enclosing side lengths stayed the same.
• When the width of the rectangle increased, the length of the rectangle decreased by half the amount.
• As the width of the rectangle increased from a very small value​​​​​​, the area of the rectangle first increased then later decreased.

  

  

  

• The maximum area of the rectangle occurred when the width was twice the length.

Solution

The railing cannot be bent or cut; whole pieces must be used.

The yard is enclosed on \(3\) sides, so

Select values for length, then calculate width and area.

Two arrangements make the greatest possible area of \(12\):

• \(2\) pieces for the parallel sides and \(6\) pieces for the other, or
• \(3\) pieces for the parallel sides and \(4\) pieces for the other.

Solution

Let \(l\) represent the length of the two parallel sides of the garden edging and \(w\) represent the width in metres.

We know

Since \(l\gt0\) and \(w\gt0\), we use \(0\) and \(10\) as placeholders for the smallest and largest possible values for \(l\) in the table.

We will choose the values for \(l\), going up by \(2\).

Then calculate the width and the area columns.

Solution

The maximum area occurs when the length is \(5\) m.

The width would be \(w=20-2\left(5\right)=10\) m.

The maximum area is \(A=5\left(10\right)=50\) m².

Solution