Figure 3 shows a flowerbed. Its shape is a quarter of a circle of radius $x$ metres with two equal rectangles attached to it along its radii. Each rectangle has length equal to $x$ metres and width equal to $y$ metres. Given that the area of the flowerbed is $4 \, m^2$, (a) show that y = \frac{16 - \pi x^2}{8x} (b) Hence show that the perimeter $P$ metres of the flowerbed is given by the equation $P = \frac{8}{x} + 2x$ (c) Use calculus to find the minimum value of $P$. (d) Find the width of each rectangle when the perimeter is a minimum. Give your answer to the nearest centimetre.

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Figure 3 shows a flowerbed - Edexcel - A-Level Maths Pure - Question 9 - 2012 - Paper 4

Question 9

Figure 3 shows a flowerbed. Its shape is a quarter of a circle of radius $x$ metres with two equal rectangles attached to it along its radii. Each rectangle has leng... show full transcript

Worked Solution & Example Answer:Figure 3 shows a flowerbed - Edexcel - A-Level Maths Pure - Question 9 - 2012 - Paper 4

Step 1

show that y = \frac{16 - \pi x^2}{8x}

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Answer

To find $y$ in terms of $x$ , we start from the area of the flowerbed:

The area of the quarter circle is given by: $\text{Area}_{circle} = \frac{1}{4} \pi x^2$

The area of the two rectangles is given by: $\text{Area}_{rectangles} = 2xy$

Thus, the total area is: $\frac{1}{4} \pi x^2 + 2xy = 4$

Rearranging this, we can find $y$ : $2xy = 4 - \frac{1}{4} \pi x^2$

Dividing both sides by $2x$ gives: $y = \frac{4 - \frac{1}{4} \pi x^2}{2x} = \frac{8 - \frac{1}{2} \pi x^2}{4x}$

This simplifies to: $y = \frac{16 - \pi x^2}{8x}$

Step 2

Hence show that the perimeter $P$ metres of the flowerbed is given by the equation $P = \frac{8}{x} + 2x$

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Answer

The perimeter $P$ of the flowerbed can be expressed as:

$P = \text{arc length} + \text{length of rectangles}$

The arc length of the quarter circle is: $\text{Arc length} = \frac{1}{4} \times 2\pi x = \frac{\pi x}{2}$

The total length of the rectangles (two of them) is: $\text{Length of rectangles} = 2x$

Thus, the perimeter can be written as: $P = \frac{\pi x}{2} + 2x$

Now, replacing $y$ in terms of $x$ , we can rearrange: Substituting the expression for $y$ : $P = \frac{\pi x}{2} + 2y = \frac{\pi x}{2} + 2(\frac{16 - \pi x^2}{8x})$

Through simplification, we get: $P = \frac{8}{x} + 2x$ .

Step 3

Use calculus to find the minimum value of $P$.

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Answer

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

$P = \frac{8}{x} + 2x$

The first derivative is: $P' = -\frac{8}{x^2} + 2$

Setting the derivative $P'$ equal to zero for critical points yields: $0 = -\frac{8}{x^2} + 2$ $\frac{8}{x^2} = 2$

Solving for $x$ , we get:

x = 2$$ Next, we should confirm that this critical point indeed provides a minimum by checking the second derivative: $$P'' = \frac{16}{x^3}$$ For $x > 0$, $P'' > 0$, confirming a local minimum at $x = 2$. Finally, substituting $x=2$ back into the equation for $P$: $$P = \frac{8}{2} + 2(2) = 4 + 4 = 8 \text{ metres}.$

Step 4

Find the width of each rectangle when the perimeter is a minimum. Give your answer to the nearest centimetre.

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Answer

Since we found $x = 2$ , we can now determine $y$ :

Using the derived formula: $y = \frac{16 - \pi (2)^2}{8(2)}$ $y = \frac{16 - 4\pi}{16}$ $y = 1 - \frac{\pi}{4}$

Calculating this gives approximately: $y \approx 1 - 0.785 = 0.215 \text{ metres}$

To find the width of each rectangle, we convert to centimetres: $\text{Width} \approx 0.215 \times 100 = 21.5 \text{ cm}$

Thus, to the nearest centimetre, the width is: $$\text{Width} \approx 22 \text{ cm}.$

Figure 3 shows a flowerbed - Edexcel - A-Level Maths Pure - Question 9 - 2012 - Paper 4

Worked Solution & Example Answer:Figure 3 shows a flowerbed - Edexcel - A-Level Maths Pure - Question 9 - 2012 - Paper 4

show that y = \frac{16 - \pi x^2}{8x}

Hence show that the perimeter $P$ metres of the flowerbed is given by the equation $P = \frac{8}{x} + 2x$

Use calculus to find the minimum value of $P$.

Find the width of each rectangle when the perimeter is a minimum. Give your answer to the nearest centimetre.

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