Figure 3 shows a flowerbed - Edexcel - A-Level Maths Pure - Question 1 - 2011 - Paper 4

The perimeter P of the flowerbed consists of the boundary of the quarter circle and the sides of the two rectangles:

$P = \text{Arc of quarter circle} + \text{Lengths of rectangles}$

The arc length is:

$\text{Arc of quarter circle} = \frac{1}{4}(2\pi x) = \frac{\pi x}{2}$

The length of the two rectangles is:

$2(\text{length}) = 2 \times x = 2x$

Thus, we can write:

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

Substituting for y from part (a) gives:

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

To minimize P, we differentiate it with respect to x:

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

Calculating the derivative:

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

Setting the derivative equal to zero to find critical points:

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

Solving for x gives:

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

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

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

For x > 0, this is positive, confirming a minimum.

Using the value of x found:

$x = 2$

Substituting x back into the equation for y from part (a):

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

Calculating y:

Using the value of ( \pi \approx 3.14 ):

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

Thus the width is approximately 0.215 m, which converts to:

$0.215 \times 100 = 21.5$ cm

Rounding to the nearest centimeter gives the width as 22 cm.