Figure 3 shows a flowerbed - Edexcel - A-Level Maths Pure - Question 2 - 2012 - Paper 4

The perimeter $P$ of the flowerbed consists of the curved section of the quarter circle and the two rectangular lengths:

$P = \frac{\pi x}{2} + 2y + 2x$ Substituting the expression for $y$ from part (a):

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

This simplifies to: $P = \frac{\pi x}{2} + \frac{32 - 2\pi x^2}{8} + 2x$

Combining terms by finding a common denominator leads to: $P = \frac{8 + 4x - \pi x^2}{4} + 2x$ $P = \frac{8}{x} + 2x$

To find the minimum of the perimeter function, we take the derivative of $P$ with respect to $x$:

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

Setting the derivative to zero to find critical points:

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

Solving for $x$ gives: $\frac{8}{x^2} = 2 \Rightarrow x^2 = 4 \Rightarrow x = 2$

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

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

Thus, the minimum value occurs at $x = 2$.

Substituting $x = 2$ into the equation for $y$:

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

Calculating the approximate value: $y \approx 2 - 1.57 \approx 0.43 \, m$

Thus, converting to centimeters: $0.43 \, m \times 100 = 43 \, cm$

Finally, rounding to the nearest centimetre gives $43 \, cm$.