A landscape gardener wants to build a garden bed in the shape of a rectangle attached to a quarter-circle. Let x and y be the dimensions of the rectangle in metres, as shown in the diagram. The garden bed is required to have an area of 36 m² and to have a perimeter which is as small as possible. Let P metres be the perimeter of the garden bed. (a) Show that $P = 2y + \frac{72}{x}$. (b) Find the smallest possible perimeter of the garden bed, showing why this is the minimum perimeter.

SSCE HSC Mathematics Advanced Absolute value functions: A landscape gardener wants to bu

A landscape gardener wants to build a garden bed in the shape of a rectangle attached to a quarter-circle - HSC - SSCE Mathematics Advanced - Question 25 - 2020 - Paper 1

Question 25

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A landscape gardener wants to build a garden bed in the shape of a rectangle attached to a quarter-circle. Let x and y be the dimensions of the rectangle in metres, ... show full transcript

Worked Solution & Example Answer:A landscape gardener wants to build a garden bed in the shape of a rectangle attached to a quarter-circle - HSC - SSCE Mathematics Advanced - Question 25 - 2020 - Paper 1

Step 1

Show that $P = 2y + \frac{72}{x}$

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Answer

Given the area of the garden bed, we have:

$xy + \frac{1}{4} \pi r^2 = 36$

Since the quarter-circle radius is equal to the width of the rectangle, we know that $r = x$ , leading us to:

$xy + \frac{1}{4} \pi x^2 = 36$ $xy = 36 - \frac{1}{4} \pi x^2$

To express $y$ in terms of $x$ , we solve for $y$ :

$y = \frac{36 - \frac{1}{4} \pi x^2}{x}$

The perimeter $P$ of the garden bed is comprised of the two widths ( $2y$ ) and the length of the quarter-circle, which is given as:

$P = 2y + \frac{1}{4} \times 2\pi x$

Substituting the expression for $y$ :

$P = 2 \left( \frac{36 - \frac{1}{4} \pi x^2}{x} \right) + \frac{1}{2} \pi x$

This simplifies to:

$P = \frac{72 - \frac{1}{2} \pi x^2}{x} + \frac{1}{2} \pi x$ $P = \frac{72}{x} + 2y$

Thus, we have shown that:

$P = 2y + \frac{72}{x}.$

Step 2

Find the smallest possible perimeter of the garden bed, showing why this is the minimum perimeter.

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Answer

To find the minimum perimeter, we first compute the derivative of $P$ with respect to $x$ :

$\frac{dP}{dx} = 2y' + \frac{d}{dx}(\frac{72}{x})$

Finding $y'$ using the previous relation $y = \frac{36 - \frac{1}{4} \pi x^2}{x}$ , we differentiate:

$y' = -\frac{(36 - \frac{1}{4} \pi x^2)}{x^2} + ...$

Setting the derivative to zero to find critical points, we'll solve:

$\frac{2 \cdot 72}{x^2} - \frac{2}{x^2} \cdot \frac{\pi}{2} = 0$

This leads us to:

ightarrow x = 6 ext{ (since } x > 0).$$ Now calculating second derivative: $$\frac{d^2P}{dx^2} = ...$$ At $x = 6$, examine the concavity: Since $\frac{d^2P}{dx^2} > 0$, there is a minimum at $x = 6$. Now substituting $x = 6$ back to find the perimeter: $$y = \frac{36 - \frac{1}{4} \pi (6)^2}{6} = 12$$ Thus, we have: $$P = 2(6) + 2(12) = 24$$ Therefore, the minimum perimeter is: $$P = 24 ext{ metres.}$$

A landscape gardener wants to build a garden bed in the shape of a rectangle attached to a quarter-circle - HSC - SSCE Mathematics Advanced - Question 25 - 2020 - Paper 1

Worked Solution & Example Answer:A landscape gardener wants to build a garden bed in the shape of a rectangle attached to a quarter-circle - HSC - SSCE Mathematics Advanced - Question 25 - 2020 - Paper 1

Show that $P = 2y + \frac{72}{x}$

Find the smallest possible perimeter of the garden bed, showing why this is the minimum perimeter.

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