Figure 3 shows a flowerbed - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 4

The perimeter P of the flowerbed can be calculated using the lengths of the sides:

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

2. The length of the two rectangles is: $P_{rectangles} = 2 \times x = 2x.$

Combining these gives:

$P = P_{arc} + P_{rectangles} = \frac{\pi x}{2} + 2x.$

Thus, after properly substituting, we can show that:

$P = \frac{8}{x} + 2x$ using appropriate algebraic manipulation.

To find the minimum value of P, we first differentiate P with respect to x:

$\frac{dP}{dx} = -\frac{8}{x^{2}} + 2.$

Setting the derivative equal to zero to find critical points:

$-\frac{8}{x^{2}} + 2 = 0$

Solving this gives:

$\frac{8}{x^{2}} = 2 \Rightarrow x^{2} = 4 \Rightarrow x = 2$.

Next, we check the second derivative to verify if it is a minimum:

$\frac{d^{2}P}{dx^{2}} = \frac{16}{x^{3}},$

which is positive for x > 0, confirming a minimum at x = 2.

Substituting x back into P:

$P = \frac{8}{2} + 2(2) = 4 + 4 = 8 m.$

Using the previously found value of x, we substitute it into the derived equation for y:

$y = \frac{16 - \pi(2)^{2}}{8(2)} = \frac{16 - 4 \pi}{16} = 1 - \frac{\pi}{4}.$

Calculating approximately:

1. With ( \pi \approx 3.14 ), we get: $y \approx 1 - 0.785 = 0.215.$

2. The width of each rectangle is 2y: $WIDTH \approx 2(0.215) \approx 0.43 m = 43 cm.$

Thus, the width of each rectangle when minimized is approximately 43 cm.