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

The perimeter P of the flowerbed consists of the arc length of the quarter circle and the lengths of the two rectangles. The arc length for a quarter circle is given by:

$Length_{arc} = \frac{1}{4} \cdot 2\pi x = \frac{\pi x}{2}$

Thus, the perimeter P can be expressed as:

$P = \frac{\pi x}{2} + 2y$

Substituting the expression for y from part (a):

$P = \frac{\pi x}{2} + 2\left(\frac{16 - \pi x^{2}}{8x}\right)$

Simplifying the equation leads to:

$P = \frac{\pi x}{2} + \frac{32 - 2\pi x^{2}}{8x} \quad = \frac{4\pi x + 32 - 2\pi x^{2}}{8x} = \frac{8}{x} + 2x$

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

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

Taking the derivative:

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

Setting the derivative equal to zero for critical points:

$-\frac{8}{x^{2}} + 2 = 0 \Rightarrow \frac{8}{x^{2}} = 2 \Rightarrow x^{2} = 4 \Rightarrow x = 2 \text{ (since } x > 0 \text{)}$

To confirm it's a minimum, check the second derivative:

$\frac{d^{2}P}{dx^{2}} = \frac{16}{x^{3}} > 0$

Since it's positive, the function is concave up and we have a minimum at x = 2. Evaluating P at this value gives:

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

Using the previously found value of x = 2, we can substitute back to find y:

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

Calculating the numerical value:

Using (\pi \approx 3.14), we get:

$1 - \frac{3.14}{4} = 1 - 0.785 = 0.215$

Thus, the width of each rectangle is approximately 0.215 m, or 21.5 cm. Rounding to the nearest centimetre gives:

22 cm.