Old MacDonald has a farm with a rock wall around it. The figure shows a portion of his farm where he plans to build two enclosures. The total length of the fence is 180 m.

1. Form an equation using the total length of fence.

There are three vertical fence segments of length $x$ and two horizontal fence segments of length $y$.

$3x+2y=180$

2. Rearrange the equation to express $y$ in terms of $x$.

$2y=180-3x$

$y=90-\frac{3}{2}x$

1. Write the total area of the two rectangular enclosures in terms of $x$ and $y$.

Each enclosure has area $xy$.

For 2 enclosures,
$A=2xy$

2. Substitute $y=90-\dfrac{3}{2}x$ and expand to form the quadratic equation for $A$ in terms of $x$.

Using $y=90-\dfrac{3}{2}x$,
$A=2x\left(90-\dfrac{3}{2}x\right)$
$A=180x-3x^2$

1. Substitute $A=2100$ into the area equation and form a quadratic equation in standard form.

From part (b),
$A=180x-3x^2.$
When $A=2100$,
$2100=180x-3x^2$
$3x^2-180x+2100=0$
$x^2-60x+700=0.$

2. Solve the quadratic equation for $x$.

Using the quadratic formula on $x^2-60x+700=0$,
$x=\frac{-(-60)\pm\sqrt{(-60)^2-4(1)(700)}}{2(1)}$
$x=\frac{60\pm\sqrt{3600-2800}}{2}$
$x=\frac{60\pm\sqrt{800}}{2}.$

3. Evaluate and round both solutions to 2 decimal places.

$x=\frac{60+\sqrt{800}}{2}=44.142\ldots$
$x=\frac{60-\sqrt{800}}{2}=15.857\ldots$
To 2 decimal places,
$x=44.14\text{ m}\quad \text{or}\quad x=15.86\text{ m}.$

1. Find the $x$-coordinate of the turning point using the midpoint of the two equal-area values from part (c).

From part (c),
$x=14.47$ or $x=45.53$ when $A=2100$.

Since these two points have the same $A$-value, the axis of symmetry is halfway between them:

$x=\frac{14.47+45.53}{2}=30$

2. Substitute the $x$-coordinate into the area equation to find the maximum area value.

Using $A=180x-3x^2$,

$A=180(30)-3(30)^2$

$A=5400-2700=2700$

Therefore, the turning point is

$(30,2700)$

1. Identify the shape of the quadratic area graph.

From part (b),
$A=180x-3x^2.$
The coefficient of $x^2$ is $-3$, which is negative.

2. Use the shape of the graph to decide whether the turning point is a maximum or minimum.

Since the coefficient of $x^2$ is negative, the parabola opens downwards. Therefore, its turning point is the highest point of the graph, so the area is a maximum.