老麦克唐纳有一个周围有石墙 of 农场。该图显示了他计划建造两个围栏的一部分农场区域。 篱笆的总长度为 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.