Mabel wants to fence off some land as an enclosure for her chickens.
The enclosure will be a rectangle with an area of 60 m^2.

1. Find the width of the rectangle in terms of $x$.

Area of rectangle $=$ length $\times$ width.

Since the area is $60\text{ m}^2$ and the length is $x\text{ m}$,
$x \times \text{width}=60$
$\text{width}=\frac{60}{x}\text{ m}$

2. Use the perimeter formula for a rectangle and simplify.

Perimeter of rectangle $=2(\text{length}+\text{width})$.

$P=2\left(x+\frac{60}{x}\right)$
$P=2x+\frac{120}{x}$

1. Evaluate $P$ when $x=2$ to find the missing table value $k$.

When $x=2$, $P=k$.

$k=2(2)+\frac{120}{2}=4+60=64$

1. List the coordinate points to be plotted, using the missing value found earlier.

From the table and part (b), $k=64$.

The points are
$(2,64),\ (4,38),\ (6,32),\ (8,31),\ (10,32),\ (12,34),\ (14,36.6).$

2. Plot the points accurately on the given grid.

On the horizontal axis use the given $x$-values $2,4,6,8,10,12,14$.

On the vertical axis plot the corresponding $P$-values $64,38,32,31,32,34,36.6$ accurately.

3. Draw the graph over the required interval.

Join the plotted points with a smooth curve, not straight line segments, for
$2\le x\le 14.$

1. Represent the condition that Mabel has only 35 m of fencing on the graph.

Since Mabel has $35$ m of fencing, she needs
$P\leq 35.$
On the graph of $P=2x+\dfrac{120}{x}$, draw the horizontal line
$P=35.$

2. Read the two intersection points and state the interval where the curve satisfies the constraint.

From the graph, the line $P=35$ meets the curve at approximately
$x=4.7 \quad \text{and} \quad x=12.8.$
The curve is at or below $P=35$ between these two values, so
$4.7\leq x\leq 12.8.$

1. Relate the perimeter to the side length when the enclosure is a square.

If the enclosure is a square, its length and width are both $x$ metres.

So its perimeter is
$P=4x.$
Therefore Mabel should draw the straight line $P=4x$ on the same graph and read its intersection with the curve.