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Rectangles Enclosed on Four Sides

Explore This 1

Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0. https://ggbm.at/kmgdbnzz#

Explore This 1 Summary

After completing the activity, you might have noticed the following:

The area of the rectangle remained constant.
Starting from $1$, as the width of the rectangle increased, the length decreased.

Starting from $1$, as the width of the rectangle increased, the perimeter decreased then later increased.
When the perimeter was a minimum, the width and the length of the rectangle were equal.

Example 1

A rectangle has an area of $12$. Is the relation between the width and the length linear or non-linear? Justify your answer using first differences and using a graph.

Solution — First Differences

To use first differences, we need a table of values. We will choose width values starting with $1$ and increase by one.

First differences are negative — this tells us that as the width increases, the length decreases.
First differences are not equal — the relation between the width and the length is non-linear.

Width, $w$	Length, $l$	First Difference in Length
$1$	$12$
$1$	$12$	$6-12=-6$
$2$	$6$	$6-12=-6$
$2$	$6$	$4-6=-2$
$3$	$4$	$4-6=-2$
$3$	$4$	$3-4=-1$
$4$	$3$	$3-4=-1$
$4$	$3$	$2.4-3=-0.6$
$5$	$2.4$	$2.4-3=-0.6$
$5$	$2.4$	$2-2.4=-0.4$
$6$	$2$	$2-2.4=-0.4$
$6$	$2$

Example 1 — Graph

A rectangle has an area of $12$. Is the relation between the width and the length linear or non-linear? Justify your answer using first differences and using a graph.

Solution — Graph

Independent variable: width, $w$

Dependent variable: length, $l$

Relation is non-linear since it is not possible to connect all our points using a single straight line.
A curve is needed to connect the points.

Example 2

A rectangle has an area of $24$. Make a table of values and a graph of the relation between the rectangle's width and perimeter. Determine the dimensions when the perimeter is a minimum.

Solution — Table of Values: Part 1

Choose width values that increase by one.
Then calculate the length and the perimeter of the rectangle.

As the width increases, the perimeter decreases and then increases.
The table suggests that the minimum perimeter occurs when the width is close to $5$.

Width, $w$	Length, $l=\dfrac{24}{w}$	Perimeter, $P=2w+2l$
$2$	$12$	$28$
$3$	$8$	$22$
$4$	$6$	$20$
$5$	$4.8$	$19.6$
$6$	$4$	$20$
$7$	$\approx3.4$	$\approx20.9$
$8$	$3$	$22$

Example 2 — Graph

A rectangle has an area of $24$. Make a table of values and a graph of the relation between the rectangle's width and perimeter. Determine the dimensions where the perimeter is a minimum.

Solution — Graph

Example 2 — Table of Values Again

A rectangle has an area of $24$. Make a table of values and a graph of the relation between the rectangle's width and perimeter. Determine the dimensions where the perimeter is a minimum.

Solution — Table of Values: Part 2

Width	Length	Perimeter
$4.7$	$\approx5.11$	$\approx19.613$
$4.8$	$5$	$19.6$
$4.9$	$\approx4.90$	$\approx19.596$
$5$	$4.8$	$19.6$
$5.1$	$\approx4.71$	$\approx19.611$

the width is $4.9$, and
the length is approximately $4.9$.

Side length would be $\sqrt{24}\approx4.8990$.
Perimeter of the square would be $4\sqrt{24}\approx19.5959$.

Paused Finished

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Minimizing the Perimeter of a Rectangle Enclosed on Four Sides

For a rectangle enclosed on four sides with a fixed area, the minimum perimeter occurs when the length and the width of the rectangle are equal.

Example 3

A rectangular vegetable garden is to have an area of $256$ m² and a fence around all four sides. If fencing material costs $$27$/m, what is the lowest possible cost for the fence?

Solution

For the lowest possible fence cost, we will need the perimeter of the garden to be a minimum.

The minimum perimeter occurs when the rectangle is a square.

Let $s$ represent the side length of the garden in metres, where $s\gt0$.

A square with side lengths s.

Since we know the area of the garden, we can write an equation and solve for $s$.

\[\begin{align*} s^2&=A\\ s^2 &= 256 \\ s &= \sqrt{256} = 16 \end{align*} \]

The minimum perimeter would be $4s=4\left(16\right)=64$ m.

The lowest possible cost for the fence is $64\left($27\right)$ or $$1728$.

Check Your Understanding 1

Part 1: null

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Width, \(w\)	Length, \(l\)	First Difference in Length
\(1\)	\(12\)
\(1\)	\(12\)	\(6-12=-6\)
\(2\)	\(6\)	\(6-12=-6\)
\(2\)	\(6\)	\(4-6=-2\)
\(3\)	\(4\)	\(4-6=-2\)
\(3\)	\(4\)	\(3-4=-1\)
\(4\)	\(3\)	\(3-4=-1\)
\(4\)	\(3\)	\(2.4-3=-0.6\)
\(5\)	\(2.4\)	\(2.4-3=-0.6\)
\(5\)	\(2.4\)	\(2-2.4=-0.4\)
\(6\)	\(2\)	\(2-2.4=-0.4\)
\(6\)	\(2\)

Width, \(w\)	Length, \(l=\dfrac{24}{w}\)	Perimeter, \(P=2w+2l\)
\(2\)	\(12\)	\(28\)
\(3\)	\(8\)	\(22\)
\(4\)	\(6\)	\(20\)
\(5\)	\(4.8\)	\(19.6\)
\(6\)	\(4\)	\(20\)
\(7\)	\(\approx3.4\)	\(\approx20.9\)
\(8\)	\(3\)	\(22\)

Width	Length	Perimeter
\(4.7\)	\(\approx5.11\)	\(\approx19.613\)
\(4.8\)	\(5\)	\(19.6\)
\(4.9\)	\(\approx4.90\)	\(\approx19.596\)
\(5\)	\(4.8\)	\(19.6\)
\(5.1\)	\(\approx4.71\)	\(\approx19.611\)

Rectangles Enclosed on Four Sides

Explore This 1

Slide Notes

Glossary

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Explore This 1 Summary

Example 1

Solution — First Differences

Example 1 — Graph

Solution — Graph

Example 2

Solution — Table of Values: Part 1

Example 2 — Graph

Solution — Graph

Example 2 — Table of Values Again

Solution — Table of Values: Part 2

Minimizing the Perimeter of a Rectangle Enclosed on Four Sides

Example 3

Solution

Check Your Understanding 1