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 find the perimeter of the sector. The perimeter ($P$) of the sector is given by:

$P = r \times \theta + 2r$

Substituting the known values:

$P = 5 \times 0.7 + 2 \times 5 = 3.5 + 10 = 13.5\ m$

Now, to find the cost of the edging strip, we multiply the perimeter by the cost per meter:

$Cost = P \times £1.80 = 13.5 \times 1.80 = £24.30$

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

Given the area of the flowerbed is to be 20 m², we start with:

$Area = \frac{1}{2} r \times \theta$

Setting this area equal to 20 gives:

$20 = \frac{1}{2} r \theta$

From this, we derive:

$\theta = \frac{40}{r}$

Now substituting back to find the perimeter:

$P = r \times \theta + 2r = r \times \frac{40}{r} + 2r = 40 + 2r$

Substituting this into the cost function:

$C = P \times £1.80 = (40 + 2r) \times £1.80 = £72 + £3.60r$

When expressing this as required:

$C = \frac{18}{5}(20 + r)$

To find the minimum cost, we start with the expression derived:

$C = \frac{72}{r} + 18$

Calculating the derivative:

$\frac{dC}{dr} = -\frac{72}{r^2} + 0$

Setting the derivative to zero to find critical points:

$\frac{dC}{dr} = 0$ leads to:

$72 = 18r \Rightarrow r = \frac{72}{18} = 4\ m$

Now, checking the second derivative:

$\frac{d^2C}{dr^2} = \frac{144}{r^3}$

Since $\frac{d^2C}{dr^2} > 0$, this indicates a minimum occurs here. Therefore, the minimum cost occurs approximately when $r \approx 4.5$.