A sector of a circle \( O A B \) with radius \( r \mathrm{~cm} \) contains an angle of \( \theta \) radians between the bounding radii. (i) Given that the perimeter of the sector is 7 cm , express \( \theta \) in terms of \( r \) and show that the area of the sector is \( \frac{1}{2} r(7-2 r) \mathrm{cm}^{2} \). (ii) Hence, or otherwise, find the maximum area of this sector as \( r \) varies.

(i) The perimeter of the sector consists of the two radii and the arc. Since the arc length is \( r\theta \) the perimeter is \[ 2r + r\theta = 7. \] Solving for \( \theta \) we have \[ r\theta = 7 - 2r \quad \Longrightarrow \quad \theta = \frac{7-2r}{r}. \] The area of a sector with angle \( \theta \) and radius \( r \) is given by \[ \text{Area} = \frac{1}{2}r^2\theta. \] Substitute the expression for \( \theta \) into the area formula: \[ \text{Area} = \frac{1}{2}r^2\left(\frac{7-2r}{r}\right) = \frac{1}{2}r(7-2r). \] (ii) The area of the sector as a function of \( r \) is \[ A(r) = \frac{1}{2}r(7-2r) = \frac{7r - 2r^2}{2}. \] This is a quadratic function in \( r \). To maximize \( A(r) \), we can differentiate with respect to \( r \) or complete the square. Differentiate \( A(r) \): \[ \frac{dA}{dr} = \frac{7 - 4r}{2}. \] Setting the derivative equal to zero gives \[ \frac{7 - 4r}{2} = 0 \quad \Longrightarrow \quad 7 - 4r = 0 \quad \Longrightarrow \quad r = \frac{7}{4}. \] Substitute \( r = \frac{7}{4} \) into the area formula: \[ A\left(\frac{7}{4}\right) = \frac{1}{2}\left(\frac{7}{4}\right)\left(7-2\left(\frac{7}{4}\right)\right). \] Simplify the expression inside the parentheses: \[ 7 - 2\left(\frac{7}{4}\right) = 7 - \frac{7}{2} = \frac{14-7}{2} = \frac{7}{2}. \] Thus, \[ A\left(\frac{7}{4}\right) = \frac{1}{2}\left(\frac{7}{4}\right)\left(\frac{7}{2}\right) = \frac{7 \cdot 7}{16} = \frac{49}{16}\ \text{cm}^2. \] The maximum area of the sector is therefore \[ \frac{49}{16}\ \text{cm}^2. \]