a) Let \( A_R \) denote the area of the rectangle and \( A_S \) denote the area of the semicircle [M1]
One correct area \( A_R = 2xh \) OR \( A_S = \frac{1}{2}\pi x^2 \) [A1]
\( A = 2xh + \frac{1}{2}\pi x^2 = x\left(2h + \frac{1}{2}\pi x\right) \) [A1] [3]

b) Attempts to find a correct expression for the total perimeter [M1]
\( (P =) 2x + 2h + \pi x \), base + 2 x height + half circumference [A1]
\( 2x + 2h + \pi x = 10 \) OR \( 2h = 10 – 2x – \pi x \) [M1]
\( h = \frac{1}{2}(10 – 2x – \pi x) \) [A1] [4]

c) \( L = 3(2xh) + \frac{1}{2}\pi x^2 \) [M1]
Substitutes \( h = \frac{1}{2}(10 – 2x – \pi x) \) into their expression for \( L \) [M1]
\( L = 6x\left(\frac{1}{2}(10 – 2x – \pi x)\right) + \frac{1}{2}\pi x^2 \) [A1]
\( = 30x – 6x^2 – 3\pi x^2 + \frac{1}{2}\pi x^2 = \left(30x – \left(6 + \frac{5\pi}{2}\right)x^2\right) \) [A1]
\( L = 30x – 6x^2 – \frac{5}{2}\pi x^2 \) [A1] [5]

d) (i) \( \frac{dL}{dx} = 30 – 12x – 5\pi x = 30 – (12 + 5\pi)x \quad (\text{accept } \frac{dL}{dx} = 30 – 27.7x) \) [M1][A1] [2]
(ii) Recognition that \( \frac{dL}{dx} = 0 \) (may be represented graphically) [M1]
\( x = 1.08272… \) [A1]
\( x = 1.08 \left(=\frac{30}{12 + 5\pi}\right) \text{ m} \) [A1]
Correct reasoning to justify a maximum [M1]
\( L \) is a quadratic (function of \( x \)) with a negative coefficient of \( x^2 \) (may be represented as a sketch indicating maximum point) OR a clearly labelled sign diagram showing the change in gradient OR \( \frac{d^2L}{dx^2} = -12 – 5\pi (= -27.7079…) < 0 \) [A1] [5]
(iii) Attempts to substitute their value of \( x \) into \( h \) [M1]
\( h = 2.21654… \) [A1]
\( h = 2.22 \left(=\frac{30 + 10\pi}{12 + 5\pi}\right) \text{(m)} \) [A1] [3]