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 sector, which consists of the arc length plus the two radius lengths:

1. Arc Length Calculation:

   • The formula for the arc length is: $L = r \theta = 5 \times 0.7 = 3.5 \text{ m}$

2. Total Perimeter Calculation:

   • The total perimeter, P: $P = L + 2r = 3.5 + 2 \times 5 = 3.5 + 10 = 13.5 \text{ m}$

3. Cost Calculation:

   • The cost, C, is the perimeter multiplied by the cost per metre (£1.80): $C = 13.5 \times 1.80 = £24.30$

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

To derive the expression for the cost, we start with the area of the flowerbed:

Given that the area (A) is 20 m², we know:

$A = \frac{1}{2} r^2 \theta \Rightarrow 20 = \frac{1}{2} r^2 \times \theta$

From the earlier calculations, we will set up the perimeter equation:

1. Perimeter Calculation:

   • The perimeter is: $P = r \theta + 2r$

2. Substituting for θ:

   • To express θ in terms of A: $\theta = \frac{20}{\frac{1}{2} r^2} = \frac{40}{r^2}$
   • Substitute this back into the perimeter formula: $P = r \cdot \frac{40}{r^2} + 2r = \frac{40}{r} + 2r$

3. Cost Expression:

   • The Cost, C: $C = P \cdot 1.80 = \left( \frac{40}{r} + 2r \right) \cdot 1.80 = 1.80 \cdot \frac{40}{r} + 3.60r$
   • Simplifying, we see: $C = \frac{72}{r} + 3.60r$
   • Now, factoring out: $C = \frac{18}{5} \left( \frac{20}{r} + r \right) \text{ where } r \text{ is the radius in metres.}$

Thus, we have shown that the cost is £C = \frac{18}{5} \left( \frac{20}{r} + r \right).

To find the minimum cost, we will differentiate the cost function:

1. Differentiate the Cost Function: $C = \frac{72}{r} + 3.60r$

   The derivative $\frac{dC}{dr}$ is given by: $\frac{dC}{dr} = -\frac{72}{r^2} + 3.60$

2. Set the Derivative to Zero:

   • To find stationary points, set $\frac{dC}{dr} = 0$: $-\frac{72}{r^2} + 3.60 = 0 \Rightarrow \frac{72}{r^2} = 3.60 \Rightarrow r^2 = \frac{72}{3.60} = 20 \Rightarrow r = \sqrt{20} \approx 4.472$

3. Second Derivative Test: $\frac{d^2C}{dr^2} = \frac{144}{r^3} \text{ which is positive for } r > 0$

   • This indicates a minimum at $r \approx 4.472$.

Thus, the minimum cost of the edging strip occurs when r ≈ 4.5.