A gardener is creating flowerbeds in the shape of sectors of circles - AQA - A-Level Maths Pure - Question 5 - 2021 - Paper 3

First, we need to calculate the perimeter of the flowerbed which consists of the arc length and the two straight edges:

1. Calculate the arc length using the formula: $P = r \theta$ Substituting the values: $P = 5 \times 0.7 = 3.5 \, m$

2. The total perimeter is given by: $Total \ Perimeter = 2r + arc \ length = 2 \times 5 + 3.5 = 10 + 3.5 = 13.5 \, m$

3. The cost of the edging strip is: $Cost = Total \ Perimeter \times Cost \ per \ metre = 13.5 \times 1.80 = 24.30 \pounds$

Therefore, the cost of the edging strip required for this flowerbed is £24.30.

To find the cost expression, we begin with the area of the flowerbed given as 20 m². The perimeter can be expressed as:

1. The area formula for a circle sector is: $Area = \frac{1}{2} r^2 \theta$ For 20 m²: $20 = \frac{1}{2} r^2 \theta \Rightarrow \theta = \frac{40}{r^2}$

2. The perimeter now includes the arc and the two radii: $P = r \theta + 2r = r \left( \frac{40}{r^2} \right) + 2r = \frac{40}{r} + 2r$

3. Substitute this perimeter back into the cost formula: $C = P \times Cost \ per \ metre = \left( \frac{40}{r} + 2r \right) \times 1.80$

4. Upon simplifying it leads to: $C = \frac{72}{r} + 3.60r = \frac{18}{5}(20 + r)$ Therefore, we show it as required.

To find the minimum cost, we apply differentiation:

1. Differentiate C: $\frac{dC}{dr} = -\frac{72}{r^2} + 3.60$

2. Set the derivative to zero for critical points: $-\frac{72}{r^2} + 3.60 = 0 \Rightarrow \frac{72}{r^2} = 3.60$ $72 = 3.60r^2 \Rightarrow r^2 = \frac{72}{3.60} = 20 \Rightarrow r = \sqrt{20} \approx 4.472$

3. Checking the second derivative: $\frac{d^2C}{dr^2} = \frac{144}{r^3} > 0$ This indicates a local minimum.

Thus, the minimum cost occurs at r ≈ 4.5.